find the limit of $\lim\limits_{x\to\infty}(1+x)^{\frac{1}{x}} $

How can I find the limit : $\lim\limits_{x\to\infty}(1+x)^{\frac{1}{x}}$

I know I can use the following statement:

$$(1+x)^{1/x}=e^{\frac 1 x\log(1+x)}$$

but I don't know how to continue from there

(I'm still not allowed to use Lhoptial if that could be used here)

Solution 1:

Observe \begin{align} \lim_{x\rightarrow \infty} \exp\left( \frac{\log(1+x)}{x}\right)= \exp\left( \lim_{x\rightarrow \infty}\frac{\log(1+x)}{x}\right). \end{align} Since \begin{align} \lim_{x\rightarrow \infty}\frac{\log(1+x)}{x} = 0 \end{align} then it follows \begin{align} \lim_{x\rightarrow \infty} \exp\left( \frac{\log(1+x)}{x}\right)=1 \end{align}

Note \begin{align} \log(1+x) \leq \sqrt{x} \end{align} when $x>0$.

Additional:

The definition of $\log(1+x)$ is defined to be \begin{align} \log(1+x) := \int^x_0 \frac{dt}{1+t} \end{align} and since \begin{align} \sqrt{x} = \int^x_0 \frac{dt}{\sqrt{t}} \end{align} then we have \begin{align} \sqrt{x}-\log(1+x) = \int^x_0 \frac{1}{\sqrt{t}}-\frac{1}{t+1}\ dt\geq 0 \end{align} since \begin{align} 1+t\geq \sqrt{t} \ \ \Leftrightarrow 1-\sqrt{t}+t = \frac{3}{4}+\left(\sqrt{t}-\frac{1}{2}\right)^2\geq 0. \end{align}