How to compute limit similar to the one of $e$. How do I compute the following limit:
$$\lim\limits_{n\to \infty} \left(1+\dfrac{1}{\sqrt{n}}\right)^{\sqrt n}
$$
I'm very certain that the above converges to $e$ but don't really know how I can show that rigorously.
 A: Hint:
\begin{gather*}
\lim _{x\rightarrow \infty }( 1+f( x))^{g( x)} =e^{l}\\
where\ l=\lim _{x\rightarrow \infty } f( x) \cdotp g( x)\\
if\ and\ only\ if\ \\
\lim _{x\rightarrow \infty } f( x) =0\\
and\ \lim _{x\rightarrow \infty } g( x) =\infty 
\end{gather*}
Can you now prove what is required using the above equation?
A: $$\lim\limits_{n\to \infty} \left(1+\dfrac{1}{\sqrt{n}}\right)^{\sqrt n}$$
Let $\sqrt{n} = x$, so this limit becomes $$\lim\limits_{x\to \infty} \left(1+\dfrac{1}{x}\right)^{x}$$
because $n=x^2$, and if $n$ approaches $\infty$, then so will $x^2$, and thus $x$ will approach $\infty$ as well.
We can easily see that the above limit is $e$, so your original limit is $e$ as well.
A: Note that for every $n$, there exists a positive natural no. $m$ (depending upon $n$) such that 
$\frac 1{m+1}\le \frac 1{\sqrt n}\lt \frac 1 {m} \implies 1+\frac 1{m+1}\le 1+\frac 1{\sqrt n}\lt 1+ \frac 1 {m}$ 
Noting that $m\lt \sqrt n\le m+1$, we have
$(1+\frac 1{m+1})^{m}\le (1+\frac 1{\sqrt n})^\sqrt n\le (1+ \frac 1 m)^{m+1}$
Note that $(1+\frac 1{m+1})^{m}=(1+\frac 1{m+1})^{m+1}(1+\frac 1{m+1})^{-1}\to e\times 1=e$
Can you finish now using Squeeze theorem?
A: In case you are allowed to use logarithm and Maclaurin series expansion:
$$
L = e^{\sqrt{n} \log(1+\frac{1}{\sqrt{n}})} \sim e^{\sqrt{n}(\frac{1}{\sqrt{n}}-\frac{1}{2n} + O(n^{-\frac{3}{2}}))} = e\cdot e^{-\frac{1}{2 \sqrt{n}}+O(\frac{1}{n})} \to_n e
$$
A: $a_n:=(1+1/√n)^{√n}; $
Assume you have shown:
$a_n$ is bounded above and increasing.
$a_n$ is convergent.
Consider the subsequence
$a_{n_k}=a_{k^2}=(1+1/k)^k;$
$\lim_{k \rightarrow \infty} a_{n_k}=e$.
A: For sake of simplicity, lets take the argument of the limit and manipulate it:
$(1+\frac{1}{\sqrt n})^{\sqrt n}$=$exp(ln((1+\frac{1}{\sqrt n})^{\sqrt n})$=$exp\frac{ln(1+\frac{1}{\sqrt n})}{n^{-1/2}}$
Then using L’hopital’s rule and the fact that the limit exists, we know that the limit of an exponential is equal to the exponential of the limit.This is what we get after some simplification:
$\frac{\frac{d}{dn}ln(1+\frac{1}{\sqrt n})}{\frac{d}{dn}n^{-1/2}}$=$\frac{\sqrt n}{\sqrt n+1}$
Therefore, because $\frac{\sqrt n}{\sqrt n}$=1~ $\frac{\sqrt n}{\sqrt n+1}$, we get that:
Limit=$e^{\lim_{n\to ∞} \frac{\sqrt n}{\sqrt n+1}}$= $e^{\lim_{n\to ∞} 1}$=$e^1$=e
There is no need for n≠0 here because this is a limit and is a removable discontinuity.
There an actual fairly rigorous proof. Please tell me if there is anything wrong!
