L'Hospital's Rule

## **Calculus Concept**

**L'Hospital's rule** is a theorem which provides a technique to evaluate limits of an indeterminate form of the type

0
0

{\displaystyle \frac{0}{0}}

 or

∞
∞

{\displaystyle \frac{\infty}{\infty} }

 (including negative infinities). The rule has four cases:

(1) If

lim

x
→
a

f
(
x
# )

lim

x
→
a

g
(
x
# )
0

{\displaystyle \lim\_{x \rightarrow a}f(x) = \lim\_{x \rightarrow a}g(x) = 0 }

 and

lim

x
→
a

f
′

(
x
)

g
′

(
x
)

{\displaystyle \lim\_{x \rightarrow a}\frac{f'(x)}{g'(x)} }

 exists, then

lim

x
→
a

f
(
x
)

g
(
x
)

=

lim

x
→
a

f
′

(
x
)

g
′

(
x
)

{\displaystyle \lim\_{x \rightarrow a}\frac{f(x)}{g(x)} = \lim\_{x \rightarrow a}\frac{f'(x)}{g'(x)}}

;

(2) If

lim

x
→
a

f
(
x
# )

lim

x
→
a

g
(
x
# )
∞

{\displaystyle \lim\_{x \rightarrow a}f(x) = \lim\_{x \rightarrow a}g(x) = \infty }

 and

lim

x
→
a

f
′

(
x
)

g
′

(
x
)

{\displaystyle \lim\_{x \rightarrow a}\frac{f'(x)}{g'(x)} }

 exists, then

lim

x
→
a

f
(
x
)

g
(
x
)

=

lim

x
→
a

f
′

(
x
)

g
′

(
x
)

{\displaystyle \lim\_{x \rightarrow a}\frac{f(x)}{g(x)} = \lim\_{x \rightarrow a}\frac{f'(x)}{g'(x)}}

;

(3) If

lim

x
→
∞

f
(
x
# )

lim

x
→
∞

g
(
x
# )
0

{\displaystyle \lim\_{x \rightarrow \infty}f(x) = \lim\_{x \rightarrow \infty}g(x) = 0 }

 and

lim

x
→
∞

f
′

(
x
)

g
′

(
x
)

{\displaystyle \lim\_{x \rightarrow \infty}\frac{f'(x)}{g'(x)} }

 exists, then

lim

x
→
∞

f
(
x
)

g
(
x
)

=

lim

x
→
∞

f
′

(
x
)

g
′

(
x
)

{\displaystyle \lim\_{x \rightarrow \infty}\frac{f(x)}{g(x)} = \lim\_{x \rightarrow \infty}\frac{f'(x)}{g'(x)}}

;

(4) If

lim

x
→
∞

f
(
x
# )

lim

x
→
∞

g
(
x
# )
∞

{\displaystyle \lim\_{x \rightarrow \infty}f(x) = \lim\_{x \rightarrow \infty}g(x) = \infty }

 and

lim

x
→
∞

f
′

(
x
)

g
′

(
x
)

{\displaystyle \lim\_{x \rightarrow \infty}\frac{f'(x)}{g'(x)} }

 exists, then

lim

x
→
∞

f
(
x
)

g
(
x
)

=

lim

x
→
∞

f
′

(
x
)

g
′

(
x
)

{\displaystyle \lim\_{x \rightarrow \infty}\frac{f(x)}{g(x)} = \lim\_{x \rightarrow \infty}\frac{f'(x)}{g'(x)}}

.

## **Problem**

Although the rule is named after the 17th-century French mathematician Guillaume de L'Hospital (1661 – 1704 AD), the theorem was actually introduced to him by the Swiss mathematician Johann Bernoulli (1667 – 1748) in 1694. L'Hospital then published this rule in his book *Analyse des Infiniment Petits pour l'Intelligence des Lignes Courbes* (*Analysis of the Infinitely Small for the Understanding of Curved Lines*) in 1696. This work was the first textbook on differential calculus!

Use L'Hospital's rule to calculate the limit

lim

x
→

0

x
x

{\displaystyle \lim\_{x \rightarrow 0^+} x^x }

.

## **Solution**

What happens if we blindly "plug in" the limit here?

lim

x
→

0

x
x

=

0
0

{\displaystyle \lim\_{x \rightarrow 0^+} x^x = 0^0 }

Turns out this is an indeterminate form but not of the type

0
0

{\displaystyle \frac{0}{0}}

 or

∞
∞

{\displaystyle \frac{\infty}{\infty} }

. However, we can try taking the logarithm. Let the limit be

L

{\displaystyle L}

# L

lim

x
→

0

x
x

{\displaystyle L = \lim\_{x \rightarrow 0^+} x^x }

then

ln
⁡

(
L
)

=

lim

x
→

0

ln
⁡

(

x
x

)

=

lim

x
→
0

x
ln
⁡

(
x
)

.

{\displaystyle \ln{(L)} = \lim\_{x \rightarrow 0^+} \ln{(x^x)} = \lim\_{x \rightarrow 0} x \ln{(x)}.}

Now force the above function into becoming a rational function

ln
⁡

(
L
)

=

lim

x
→

0

ln
⁡

(
x
)

1

/

x

{\displaystyle \ln{(L)} = \lim\_{x \rightarrow 0^+} \frac{\ln{(x)}}{1/x} }

and apply L'Hospital's rule

ln
⁡

(
L
)

=

lim

x
→

0

1

/

x

−
1

/

x
2

{\displaystyle \ln{(L)} = \lim\_{x \rightarrow 0^+} \frac{1/x}{-1/x^2} }

ln
⁡

(
L
)

=

lim

x
→

0

−
x

{\displaystyle \ln{(L)} = \lim\_{x \rightarrow 0^+} -x }

ln
⁡

(
L
)

=

{\displaystyle \ln{(L)} = 0. }

However, we computed the logarithm of the limit, so the limit is actually

# L

e
0

=

{\displaystyle L = e^0 = 1. }

The graph shows it (Figure 2).