Q:

How to show this $\lim_{n\rightarrow\infty} {\sqrt[n]{1+n^2}} =1$

$$ \lim_{n\rightarrow\infty} {\sqrt[n]{1+n^2}} =1$$ 
I know that this is true for n but for this expression I dont know. 

A:

Hint:
$\sqrt[n]{n^2}<\sqrt[n]{1+n^2}<\sqrt[n]{(n+1)^2}$
$\displaystyle\lim_{n \to \infty}\sqrt[n]{n^2}=1=\displaystyle\lim_{n \to \infty}\sqrt[n]{(n+1)^2}$

A:

Consider the function
$$f(x)=(x^2+1)^{1/x}=e^{\frac{\ln(x^2+1)}x},\quad x>0$$
Then
$$\lim_{x\to\infty}\frac{\ln(x^2+1)}x\stackrel*=\lim_{x\to\infty}\frac2{x^2+1}=0$$
where l'Hopital's rule has been applied at (*). 
Since the exponential function is continuous,
$$\lim_{x\to\infty}f(x)=e^0=1$$

