TITLE: verifying $\lim\limits_{x \to \infty} (1+\frac{1}{\sqrt{x}})^x$
QUESTION [5 upvotes]: What is the limit of $\lim\limits_{x \to \infty} (1+\frac{1}{\sqrt{x}})^x$ ?
I tried to solve it but I am not sure if it is appropriate to solve it this way.
$(1+\frac{1}{\sqrt{x}})^x =\exp\Bigl({x\times\ln\bigl(1+\frac{1}{\sqrt{x}}\bigr)\Bigr)}\tag{*}$
Let $\,t=\frac {1}{\sqrt{x}} $
$\,t=\frac {1}{\sqrt{x}} \Rightarrow t^2=\frac {1}{x}\Rightarrow \frac {1}{t^2}=x $
By substituting in $(*) $ we have
$\exp\Bigl(x\ln\bigl(1+\frac{1}{\sqrt{x}}\bigr)\Bigr)=\exp\Bigl(\frac{1}{t^2}\ln(1+t)\Bigr) $
as $\quad x\rightarrow \infty ,\quad \frac{1}{\sqrt{x}}\rightarrow 0,\quad$ so $t\rightarrow 0 $
$\lim\limits_{x \to \infty} (1+\frac{1}{\sqrt{x}})^x = \lim\limits_{t \to 0} \exp\bigl(\frac{1}{t^2}\ln(1+t)\bigr)$
as $t\neq 0 \,$ we can divide and multiply by $t$:
\begin{align}
\lim_{t \to 0} 
\exp\Bigl(\frac{1}{t^2}\times \ln(1+t)\Bigr)&=\lim_{t \to 0} \exp\Bigl(\frac{1}{t^2}\times \ln(1+t)\times \frac{t}{t}\Bigr)\\
&=\lim_{t \to 0} \exp\Bigl(\frac{1}{t}\times \frac {\ln(1+t)}{t}\Bigr)
\end{align}
using L’Hospital’s rule, $\,\,\lim\limits_{t \to 0} \frac{\ln(1+t)}{t}=1$
$\lim\limits_{t \to 0}\exp\Bigl(\frac{1}{t}\times \frac {\ln(1+t)}{t}\Bigr)=\infty$

REPLY [9 votes]: You can show the limit much quicker by bounding your expression from below using the Bernoulli inequality :
$$\left(1+\frac{1}{\sqrt{x}}\right)^x\geq 1+x\cdot\frac{1}{\sqrt{x}} =1+\sqrt x\stackrel{x\to +\infty}{\longrightarrow}+\infty$$