Trong giai tich, Quy tac l'Hopital (cach viet khac l'Hospital, tieng Phap: [lopital], phat am nhu Lo-pi-tan), cung đuoc goi la quy tac Bernoulli, la quy tac su dung đao ham đe tinh toan cac gioi han co dang vo đinh. Ung dung cua quy tac nay la đua dang vo đinh tro thanh dang huu han, cho phep tinh toan gioi han mot cach de dang. Quy tac nay đuoc đat theo ten cua nha toan hoc nguoi Phap Guillaume de l'Hopital. Ong đa phat bieu quy tac nay trong cuon sach Analyse des Infiniment Petits pour l'Intelligence des Lignes Courbes (1696) cua minh - cuon sach đau tien ve vi phan. Tuy nhien, cong thuc nay đuoc cho la do nha toan hoc nguoi Thuy Si Johann Bernoulli phat hien.
Đinh ly Stolz–Cesaro la mot ket qua tuong tu ve gioi han cua cac day, nhung no chi su dung sai phan huu han thay vi đao ham.
Dang đon gian nhat cua quy tac l'Hopital đuoc phat bieu nhu sau: Cho hai ham so ƒ va g:
Neu 
  
    
      
        
          lim
          
            x
            →
            c
          
        
        f
        (
        x
        )
        =
        
          lim
          
            x
            →
            c
          
        
        g
        (
        x
        )
        =
        0
      
    
    {\textstyle \lim _{x\to c}f(x)=\lim _{x\to c}g(x)=0}
  

hoac 
  
    
      
        ±
        ∞
      
    
    {\textstyle \pm \infty }
  
 va 
  
    
      
        
          lim
          
            x
            →
            c
          
        
        
          
            
              
                f
                ′
              
              (
              x
              )
            
            
              
                g
                ′
              
              (
              x
              )
            
          
        
      
    
    {\textstyle \lim _{x\to c}{\frac {f'(x)}{g'(x)}}}
  
 ton tai, thi

  
    
      
        
          lim
          
            x
            →
            c
          
        
        
          
            
              f
              (
              x
              )
            
            
              g
              (
              x
              )
            
          
        
        =
        
          lim
          
            x
            →
            c
          
        
        
          
            
              
                f
                ′
              
              (
              x
              )
            
            
              
                g
                ′
              
              (
              x
              )
            
          
        
      
    
    {\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}}
  
.
Viec lay đao ham cua tu so va mau so thuong lam đon gian thuong so, hoac lam khu dang vo đinh.


Dang tong quat Dang chung cua quy tac l'Hopital bao gom nhieu truong hop khac. Gia su c va L la cac so thuoc tap so thuc mo rong (tuc la bao gom tap so thuc va hai gia tri duong vo cung va am vo cung). Neu

  
    
      
        
          lim
          
            x
            →
            c
          
        
        
          f
          (
          x
          )
        
        =
        
          lim
          
            x
            →
            c
          
        
        g
        (
        x
        )
        =
        0
      
    
    {\displaystyle \lim _{x\to c}{f(x)}=\lim _{x\to c}g(x)=0}
  

hoac

  
    
      
        
          lim
          
            x
            →
            c
          
        
        
          f
          (
          x
          )
        
        =
        ±
        
          lim
          
            x
            →
            c
          
        
        
          g
          (
          x
          )
        
        =
        ±
        ∞
      
    
    {\displaystyle \lim _{x\to c}{f(x)}=\pm \lim _{x\to c}{g(x)}=\pm \infty }
  
.
Va gia su

  
    
      
        
          lim
          
            x
            →
            c
          
        
        
          
            
              
                f
                ′
              
              (
              x
              )
            
            
              
                g
                ′
              
              (
              x
              )
            
          
        
        =
        L
      
    
    {\displaystyle \lim _{x\to c}{\frac {f'(x)}{g'(x)}}=L}
  
.
thi

  
    
      
        
          lim
          
            x
            →
            c
          
        
        
          
            
              f
              (
              x
              )
            
            
              g
              (
              x
              )
            
          
        
        =
        L
      
    
    {\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=L}
  
.
Quy tac nay van đung đoi voi gioi han mot ben.


Đieu kien ton tai gioi han Đieu kien gioi han

  
    
      
        
          lim
          
            x
            →
            c
          
        
        
          
            
              
                f
                ′
              
              (
              x
              )
            
            
              
                g
                ′
              
              (
              x
              )
            
          
        
      
    
    {\displaystyle \lim _{x\to c}{\frac {f'(x)}{g'(x)}}}
  

ton tai la can thiet. Thuc hien phep lay vi phan cua mot gioi han dang vo đinh co the dan đen mot gioi han khong ton tai, va neu đieu nay xay ra thi quy tac l'Hopital se khong đung. Vi du, neu ƒ(x) = x + sin(x) va g(x) = x, thi

  
    
      
        
          lim
          
            x
            →
            ∞
          
        
        
          
            
              
                f
                ′
              
              (
              x
              )
            
            
              
                g
                ′
              
              (
              x
              )
            
          
        
        =
        
          lim
          
            x
            →
            ∞
          
        
        
          
            
              1
              +
              cos
              ⁡
              x
            
            1
          
        
      
    
    {\displaystyle \lim _{x\to \infty }{\frac {f'(x)}{g'(x)}}=\lim _{x\to \infty }{\frac {1+\cos x}{1}}}
  
,
khong ton tai, trong khi

  
    
      
        
          lim
          
            x
            →
            ∞
          
        
        
          
            
              f
              (
              x
              )
            
            
              g
              (
              x
              )
            
          
        
        =
        
          lim
          
            x
            →
            ∞
          
        
        
          (
          
            1
            +
            
              
                
                  sin
                  ⁡
                  x
                
                x
              
            
          
          )
        
        =
        1
      
    
    {\displaystyle \lim _{x\to \infty }{\frac {f(x)}{g(x)}}=\lim _{x\to \infty }\left(1+{\frac {\sin x}{x}}\right)=1}
  
.


Vi du Đay la mot vi du lien quan đen ham sinc va dang vo đinh 0/0:

  
    
      
        
          
            
              
                
                  lim
                  
                    x
                    →
                    0
                  
                
                sinc
                ⁡
                (
                x
                )
              
              
                
                =
                
                  lim
                  
                    x
                    →
                    0
                  
                
                
                  
                    
                      sin
                      ⁡
                      π
                      x
                    
                    
                      π
                      x
                    
                  
                
              
            
            
              
              
                
                =
                
                  lim
                  
                    y
                    →
                    0
                  
                
                
                  
                    
                      sin
                      ⁡
                      y
                    
                    y
                  
                
              
            
            
              
              
                
                =
                
                  lim
                  
                    y
                    →
                    0
                  
                
                
                  
                    
                      cos
                      ⁡
                      y
                    
                    1
                  
                
              
            
            
              
              
                
                =
                1
              
            
          
        
      
    
    {\displaystyle {\begin{aligned}\lim _{x\to 0}\operatorname {sinc} (x)&=\lim _{x\to 0}{\frac {\sin \pi x}{\pi x}}\\&=\lim _{y\to 0}{\frac {\sin y}{y}}\\&=\lim _{y\to 0}{\frac {\cos y}{1}}\\&=1\end{aligned}}}
  
.
Nhu vay, gioi han tren la đinh nghia cua đao ham ham so sin tai 0.
Đay la mot vi du phuc tap hon ve dang 0/0: sau khi ap dung quy tac l'Hopital van dan toi mot dang vo đinh. Trong nhung truong hop nay, ta co the ap dung quy tac l’Hopital nhieu lan đe tinh gioi han:

  
    
      
        
          
            
              
                
                  lim
                  
                    x
                    →
                    0
                  
                
                
                  
                    
                      2
                      sin
                      ⁡
                      x
                      −
                      sin
                      ⁡
                      2
                      x
                    
                    
                      x
                      −
                      sin
                      ⁡
                      x
                    
                  
                
              
              
                
                =
                
                  lim
                  
                    x
                    →
                    0
                  
                
                
                  
                    
                      2
                      cos
                      ⁡
                      x
                      −
                      2
                      cos
                      ⁡
                      2
                      x
                    
                    
                      1
                      −
                      cos
                      ⁡
                      x
                    
                  
                
              
            
            
              
              
                
                =
                
                  lim
                  
                    x
                    →
                    0
                  
                
                
                  
                    
                      −
                      2
                      sin
                      ⁡
                      x
                      +
                      4
                      sin
                      ⁡
                      2
                      x
                    
                    
                      sin
                      ⁡
                      x
                    
                  
                
              
            
            
              
              
                
                =
                
                  lim
                  
                    x
                    →
                    0
                  
                
                
                  
                    
                      −
                      2
                      cos
                      ⁡
                      x
                      +
                      8
                      cos
                      ⁡
                      2
                      x
                    
                    
                      cos
                      ⁡
                      x
                    
                  
                
              
            
            
              
              
                
                =
                
                  
                    
                      −
                      2
                      +
                      8
                    
                    1
                  
                
                =
                6
              
            
          
        
      
    
    {\displaystyle {\begin{aligned}\lim _{x\to 0}{\frac {2\sin x-\sin 2x}{x-\sin x}}&=\lim _{x\to 0}{\frac {2\cos x-2\cos 2x}{1-\cos x}}\\&=\lim _{x\to 0}{\frac {-2\sin x+4\sin 2x}{\sin x}}\\&=\lim _{x\to 0}{\frac {-2\cos x+8\cos 2x}{\cos x}}\\&={\frac {-2+8}{1}}=6\end{aligned}}}
  
.
Vi du sau cung ve dang vo đinh 0/0. Gia su b > 0. Khi đo

  
    
      
        
          lim
          
            x
            →
            0
          
        
        
          
            
              
                b
                
                  x
                
              
              −
              1
            
            x
          
        
        =
        
          lim
          
            x
            →
            0
          
        
        
          
            
              
                b
                
                  x
                
              
              ln
              ⁡
              b
            
            1
          
        
        =
        ln
        ⁡
        b
        
          lim
          
            x
            →
            0
          
        
        
          
            b
            
              x
            
          
        
        =
        ln
        ⁡
        b
      
    
    {\displaystyle \lim _{x\to 0}{\frac {b^{x}-1}{x}}=\lim _{x\to 0}{\frac {b^{x}\ln b}{1}}=\ln b\lim _{x\to 0}{b^{x}}=\ln b}
  
.
Mot vi du khac ve dang 0/0:

  
    
      
        
          lim
          
            x
            →
            0
          
        
        
          
            
              
                e
                
                  x
                
              
              −
              1
              −
              x
            
            
              x
              
                2
              
            
          
        
        =
        
          lim
          
            x
            →
            0
          
        
        
          
            
              
                e
                
                  x
                
              
              −
              1
            
            
              2
              x
            
          
        
        =
        
          lim
          
            x
            →
            0
          
        
        
          
            
              e
              
                x
              
            
            2
          
        
        =
        
          
            1
            2
          
        
      
    
    {\displaystyle \lim _{x\to 0}{\frac {e^{x}-1-x}{x^{2}}}=\lim _{x\to 0}{\frac {e^{x}-1}{2x}}=\lim _{x\to 0}{\frac {e^{x}}{2}}={\frac {1}{2}}}
  
.
Vi du sau lien quan đen dang ∞/∞. Gia su n la mot so nguyen duong. Khi đo

  
    
      
        
          lim
          
            x
            →
            ∞
          
        
        
          x
          
            n
          
        
        
          e
          
            −
            x
          
        
        =
        
          lim
          
            x
            →
            ∞
          
        
        
          
            
              x
              
                n
              
            
            
              e
              
                x
              
            
          
        
        =
        
          lim
          
            x
            →
            ∞
          
        
        
          
            
              n
              
                x
                
                  n
                  −
                  1
                
              
            
            
              e
              
                x
              
            
          
        
        =
        n
        
          lim
          
            x
            →
            ∞
          
        
        
          
            
              x
              
                n
                −
                1
              
            
            
              e
              
                x
              
            
          
        
      
    
    {\displaystyle \lim _{x\to \infty }x^{n}e^{-x}=\lim _{x\to \infty }{\frac {x^{n}}{e^{x}}}=\lim _{x\to \infty }{\frac {nx^{n-1}}{e^{x}}}=n\lim _{x\to \infty }{\frac {x^{n-1}}{e^{x}}}}
  
.
Sau đo ap dung lien tiep quy tac l'Hopital cho đen khi so mu la 0, ta ket luan gioi han bang 0.
Mot vi du khac ve dang ∞/∞:

  
    
      
        
          lim
          
            x
            →
            
              0
              
                +
              
            
          
        
        x
        ln
        ⁡
        x
        =
        
          lim
          
            x
            →
            
              0
              
                +
              
            
          
        
        
          
            
              ln
              ⁡
              x
            
            
              1
              x
            
          
        
        =
        
          lim
          
            x
            →
            
              0
              
                +
              
            
          
        
        
          
            
              1
              
                /
              
              x
            
            
              −
              
                
                  1
                  
                    x
                    
                      2
                    
                  
                
              
            
          
        
        =
        
          lim
          
            x
            →
            
              0
              
                +
              
            
          
        
        −
        x
        =
        0
      
    
    {\displaystyle \lim _{x\to 0^{+}}x\ln x=\lim _{x\to 0^{+}}{\frac {\ln x}{\frac {1}{x}}}=\lim _{x\to 0^{+}}{\frac {1/x}{-{\frac {1}{x^{2}}}}}=\lim _{x\to 0^{+}}-x=0}
  
.
Nguoi ta cung su dung quy tac l'Hopital đe chung minh đinh ly sau: Neu f'' lien tuc tai x thi

  
    
      
        
          
            
              
                
                  lim
                  
                    h
                    →
                    0
                  
                
                
                  
                    
                      f
                      (
                      x
                      +
                      h
                      )
                      +
                      f
                      (
                      x
                      −
                      h
                      )
                      −
                      2
                      f
                      (
                      x
                      )
                    
                    
                      h
                      
                        2
                      
                    
                  
                
              
              
                
                =
                
                  lim
                  
                    h
                    →
                    0
                  
                
                
                  
                    
                      
                        f
                        ′
                      
                      (
                      x
                      +
                      h
                      )
                      −
                      
                        f
                        ′
                      
                      (
                      x
                      −
                      h
                      )
                    
                    
                      2
                      h
                    
                  
                
              
            
            
              
              
                
                =
                
                  f
                  ″
                
                (
                x
                )
              
            
          
        
      
    
    {\displaystyle {\begin{aligned}\lim _{h\to 0}{\frac {f(x+h)+f(x-h)-2f(x)}{h^{2}}}&=\lim _{h\to 0}{\frac {f'(x+h)-f'(x-h)}{2h}}\\&=f''(x)\end{aligned}}}
  
.
Đoi khi quy tac L'Hopital đuoc su dung mot cach kheo leo nhu sau: Cho 
  
    
      
        f
        (
        x
        )
        +
        
          f
          ′
        
        (
        x
        )
      
    
    {\displaystyle f(x)+f'(x)}
  
 hoi tu khi 
  
    
      
        x
        →
        ∞
      
    
    {\displaystyle x\to \infty }
  
. Khi đo:

  
    
      
        
          lim
          
            x
            →
            ∞
          
        
        f
        (
        x
        )
        =
        
          lim
          
            x
            →
            ∞
          
        
        
          
            
              
                e
                
                  x
                
              
              f
              (
              x
              )
            
            
              e
              
                x
              
            
          
        
        =
        
          lim
          
            x
            →
            ∞
          
        
        
          
            
              
                e
                
                  x
                
              
              (
              f
              (
              x
              )
              +
              
                f
                ′
              
              (
              x
              )
              )
            
            
              e
              
                x
              
            
          
        
        =
        
          lim
          
            x
            →
            ∞
          
        
        (
        f
        (
        x
        )
        +
        
          f
          ′
        
        (
        x
        )
        )
      
    
    {\displaystyle \lim _{x\to \infty }f(x)=\lim _{x\to \infty }{\frac {e^{x}f(x)}{e^{x}}}=\lim _{x\to \infty }{\frac {e^{x}(f(x)+f'(x))}{e^{x}}}=\lim _{x\to \infty }(f(x)+f'(x))}
  

va do đo 
  
    
      
        
          lim
          
            x
            →
            ∞
          
        
        f
        (
        x
        )
      
    
    {\displaystyle \lim _{x\to \infty }f(x)}
  
 ton tai va 
  
    
      
        
          lim
          
            x
            →
            ∞
          
        
        
          f
          ′
        
        (
        x
        )
        =
        0
      
    
    {\displaystyle \lim _{x\to \infty }f'(x)=0}
  
.


Cac dang vo đinh khac Cac dang vo đinh khac, bao gom 1∞, 00, ∞0, 0.∞, va ∞ − ∞, đoi luc co the tinh đuoc dua vao quy tac l'Hopital. Vi du, đe tinh gioi han dang ∞ − ∞, chuyen hieu cua hai ham so thanh mot thuong hai ham so:

  
    
      
        
          
            
              
                
                  lim
                  
                    x
                    →
                    1
                  
                
                
                  (
                  
                    
                      
                        x
                        
                          x
                          −
                          1
                        
                      
                    
                    −
                    
                      
                        1
                        
                          ln
                          ⁡
                          x
                        
                      
                    
                  
                  )
                
              
              
                
                =
                
                  lim
                  
                    x
                    →
                    1
                  
                
                
                  
                    
                      x
                      ln
                      ⁡
                      x
                      −
                      x
                      +
                      1
                    
                    
                      (
                      x
                      −
                      1
                      )
                      ln
                      ⁡
                      x
                    
                  
                
                =
                
                  lim
                  
                    x
                    →
                    1
                  
                
                
                  
                    
                      ln
                      ⁡
                      x
                    
                    
                      
                        
                          
                            x
                            −
                            1
                          
                          x
                        
                      
                      +
                      ln
                      ⁡
                      x
                    
                  
                
                
                (
                1
                )
              
            
            
              
              
                
                =
                
                  lim
                  
                    x
                    →
                    1
                  
                
                
                  
                    
                      x
                      ln
                      ⁡
                      x
                    
                    
                      x
                      −
                      1
                      +
                      x
                      ln
                      ⁡
                      x
                    
                  
                
                =
                
                  lim
                  
                    x
                    →
                    1
                  
                
                
                  
                    
                      1
                      +
                      ln
                      ⁡
                      x
                    
                    
                      1
                      +
                      1
                      +
                      ln
                      ⁡
                      x
                    
                  
                
                
                (
                2
                )
              
            
            
              
              
                
                =
                
                  lim
                  
                    x
                    →
                    1
                  
                
                
                  
                    
                      1
                      +
                      ln
                      ⁡
                      x
                    
                    
                      2
                      +
                      ln
                      ⁡
                      x
                    
                  
                
                =
                
                  
                    1
                    2
                  
                
              
            
          
        
      
    
    {\displaystyle {\begin{aligned}\lim _{x\to 1}\left({\frac {x}{x-1}}-{\frac {1}{\ln x}}\right)&=\lim _{x\to 1}{\frac {x\ln x-x+1}{(x-1)\ln x}}=\lim _{x\to 1}{\frac {\ln x}{{\frac {x-1}{x}}+\ln x}}\quad (1)\\&=\lim _{x\to 1}{\frac {x\ln x}{x-1+x\ln x}}=\lim _{x\to 1}{\frac {1+\ln x}{1+1+\ln x}}\quad (2)\\&=\lim _{x\to 1}{\frac {1+\ln x}{2+\ln x}}={\frac {1}{2}}\end{aligned}}}
  
.
O tren quy tac l'Hopital đa đuoc ap dung o cac buoc (1) va (2).
Quy tac l'Hopital co the đuoc dung đoi voi dang vo đinh lien quan đen so mu bang cach su dung phep tinh logarit đe chuyen so mu xuong duoi. Đay la mot vi du ve dang 00:

  
    
      
        
          lim
          
            x
            →
            
              0
              
                +
              
            
          
        
        
          x
          
            x
          
        
        =
        
          lim
          
            x
            →
            
              0
              
                +
              
            
          
        
        
          e
          
            ln
            ⁡
            
              x
              
                x
              
            
          
        
        =
        
          lim
          
            x
            →
            
              0
              
                +
              
            
          
        
        
          e
          
            x
            ln
            ⁡
            x
          
        
        =
        
          e
          
            
              lim
              
                x
                →
                
                  0
                  
                    +
                  
                
              
            
            x
            ln
            ⁡
            x
          
        
      
    
    {\displaystyle \lim _{x\to 0^{+}}x^{x}=\lim _{x\to 0^{+}}e^{\ln x^{x}}=\lim _{x\to 0^{+}}e^{x\ln x}=e^{\lim _{x\to 0^{+}}x\ln x}}
  
.
Ta đuoc phep chuyen gioi han vao trong ham so mu vi ham mu la ham lien tuc. So mu x đa đuoc "chuyen xuong duoi". Gioi han 
  
    
      
        
          
            lim
            
              x
              →
              
                0
                
                  +
                
              
            
          
          x
          ln
          ⁡
          x
        
      
    
    {\displaystyle \scriptstyle \lim _{x\to 0^{+}}x\ln x}
  
 thuoc dang vo đinh 0•(−∞), nhung nhu ta thay o tren, quy tac l'Hopital co the đuoc dung đe xac đinh gioi han nay:

  
    
      
        
          lim
          
            x
            →
            
              0
              
                +
              
            
          
        
        x
        ln
        ⁡
        x
        =
        0
      
    
    {\displaystyle \lim _{x\to 0^{+}}x\ln x=0}
  
.
Do đo

  
    
      
        
          lim
          
            x
            →
            
              0
              
                +
              
            
          
        
        
          x
          
            x
          
        
        =
        
          e
          
            0
          
        
        =
        1
      
    
    {\displaystyle \lim _{x\to 0^{+}}x^{x}=e^{0}=1}
  
.


Truong hop f va g kha vi tai c Chung minh cua quy tac l'Hopital rat đon gian trong truong hop ƒ va g kha vi tai điem c. Đay khong phai la chung minh cua quy tac l'Hopital tong quat.
Cho hai ham so  ƒ va g lien tuc va kha vi tai c, ƒ(c) = g(c) = 0, va g′(c) = 0. Khi đo

  
    
      
        
          lim
          
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    {\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f(x)-f(c)}{g(x)-g(c)}}=\lim _{x\to c}{\frac {\frac {f(x)-f(c)}{x-c}}{\frac {g(x)-g(c)}{x-c}}}={\frac {\lim _{x\to c}{\frac {f(x)-f(c)}{x-c}}}{\lim _{x\to c}{\frac {g(x)-g(c)}{x-c}}}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}}
  
.
(chu y la ƒ(c) = g(c) = 0). Đieu nay bat nguon tu quy tac tinh gioi han cua thuong va đinh nghia đao ham.
Đieu nay goi mot cach chung minh cho quy tac l'Hopital tong quat, khong yeu cau hai ham ƒ va g phai kha vi tai điem  c. Xem chung minh ben duoi.


Cach hieu hinh hoc Cho đuong cong trong mat phang, trong đo truc Ox đuoc cho boi g(t) va truc Oy cho boi ƒ(t) – vi du

  
    
      
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    {\displaystyle t\mapsto [g(t),f(t)]}
  
.
Gia su ƒ(c) = g(c) = 0. Gioi han cua ti so ƒ(t)/g(t) khi t → c la slope cua tiep tuyen đen đuong cong tai điem [0, 0]. Tiep tuyen cua đuong cong tai điem t đuoc cho boi 
  
    
      
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    {\displaystyle [g'(t),f'(t)]}
  
. Quy tac l'Hopital noi rang slope cua tiep tuyen tai 0 la gioi han cua cac slope cua cac tiep tuyen tai cac điem "rat gan" 0.


Chung minh cua quy tac l'Hopital Mot chung minh pho bien cua quy tac l'Hopital la su dung đinh ly gia tri trung gian Cauchy. Chung minh quy tac l'Hopital co mot so điem khac nhau trong cac truong hop khac nhau nhu: c va L huu han hay vo han, ƒ va g hoi tu ve 0 hay ve vo cung, gioi han la mot ben hay hai ben. Tuy nhien, tat ca chung đeu dua theo hai truong hop chinh sau:


= 0/0 =
Gia su c va L la cac so thuc va ƒ va g hoi tu ve 0.
Truoc het, ta co ƒ(c) = g(c) = 0. Vi the ƒ va g lien tuc tai c, nhung khong thay đoi gioi han (vi theo đinh nghia, gioi han khong phu thuoc vao gia tri ham tai điem c). Vi 
  
    
      
        
          
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    {\displaystyle \scriptstyle \lim _{x\to c}{\frac {f'(x)}{g'(x)}}}
  
 ton tai nen co mot khoang (c − δ, c + δ) ma voi moi x thuoc khoang, voi truong hop ngoai le x = c, ca 
  
    
      
        
          
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    {\displaystyle \scriptstyle f'(x)}
  
 va 
  
    
      
        
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    {\displaystyle g'(x)}
  
 ton tai va  
  
    
      
        
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    {\displaystyle g'(x)}
  
 khac 0.
Neu x nam trong khoang (c, c + δ), thi theo đinh ly gia tri trung gian va đinh ly gia tri trung gian Cauchy đeu ap dung đung voi khoang [c, x] (va tuong tu trong truong hop x thuoc khoang (c − δ, c)). Đinh ly gia tri trung gian noi rang g(x) khac 0 (vi neu nguoc lai thi ton tai y trong khoang (c, x) ma g'(y) = 0). Tu đinh ly gia tri trung gian Cauchy, ta suy ra co mot ξx thuoc (c, x) thoa man

  
    
      
        
          
            
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    {\displaystyle {\frac {f(x)}{g(x)}}={\frac {f'(\xi _{x})}{g'(\xi _{x})}}}
  
.
Neu x tien đen c, thi ξx tien toi c (theo nguyen ly kep). Do 
  
    
      
        
          
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    {\displaystyle \scriptstyle \lim _{x\to c}{\frac {f'(x)}{g'(x)}}}
  
 ton tai, suy ra

  
    
      
        
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    {\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(\xi _{x})}{g'(\xi _{x})}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}}
  
.


= ∞/∞ =
Gia su L la mot so huu han, c la mot so huu han duong, ƒ va g hoi tu ve +∞.
Voi moi ε > 0, ton tai mot so m sao cho

  
    
      
        
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    {\displaystyle \left|{\frac {f'(x)}{g'(x)}}-L\right|<\varepsilon }
  
 (voi 
  
    
      
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    {\displaystyle x\geq m}
  
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Theo đinh ly gia tri trung gian, neu x > m, thi g(x) = g(m) (vi neu nguoc lai thi ton tai y trong khoang (m, x) sao cho 
  
    
      
        
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    {\displaystyle g'(y)=0}
  
). Ap dung đinh ly gia tri trung gian Cauchy cho khoang [m, x], ta co

  
    
      
        
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    {\displaystyle \left|{\frac {f(x)-f(m)}{g(x)-g(m)}}-L\right|<\varepsilon }
  
 (voi 
  
    
      
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    {\displaystyle x>m}
  
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Vi ƒ hoi tu ve +∞ nen neu x đu lon, ta co ƒ(x) = ƒ(m). Viet

  
    
      
        
          
            
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    {\displaystyle {\frac {f(x)}{g(x)}}={\frac {f(x)-f(m)}{g(x)-g(m)}}\cdot {\frac {f(x)}{f(x)-f(m)}}\cdot {\frac {g(x)-g(m)}{g(x)}}}
  
.
Khi đo,

  
    
      
        
          
            
              
              
                
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    {\displaystyle {\begin{aligned}&\left|{\frac {f(x)-f(m)}{g(x)-g(m)}}\cdot {\frac {f(x)}{f(x)-f(m)}}\cdot {\frac {g(x)-g(m)}{g(x)}}-{\frac {f(x)-f(m)}{g(x)-g(m)}}\right|\\&\quad \leq \left|{\frac {f(x)-f(m)}{g(x)-g(m)}}\right|\left|{\frac {f(x)}{f(x)-f(m)}}\cdot {\frac {g(x)-g(m)}{g(x)}}-1\right|\\&\quad <(|L|+\varepsilon )\left|{\frac {f(x)}{f(x)-f(m)}}\cdot {\frac {g(x)-g(m)}{g(x)}}-1\right|\end{aligned}}}
  
.
Voi x đu lon, cai nay nho hon ε va do đo

  
    
      
        
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    {\displaystyle \left|{\frac {f(x)}{g(x)}}-L\right|<2\varepsilon }
  
. *
(*) Chu y: Co mot so buoc bi bo qua.


Ghi chu Chu thich Lien ket ngoai Quy tac l'Hopital's tai PlanetMath. Luu tru ngay 30 thang 10 nam 2008 tai Wayback Machine