Show $\lim\limits_{n\to\infty} \sqrt[n]{n^e+e^n}=e$ Why is $\lim\limits_{n\to\infty} \sqrt[n]{n^e+e^n}$ = $e$? I couldn't get this result.
 A: $$ \large (n^e + e^n)^{\frac{1}{n}} = e \left ( 1 + \frac {n^e}{e^n} \right) ^{\frac 1 n} $$
$$ e \Large \left ( 1 + \frac {n^e}{e^n} \right) ^{\frac 1 n}  = e \left ( \underbrace { \left ( 1 + \frac {1}{e^{n - e \log n}} \right) ^{e^{n - e \log n}} }_{e}\right)^{ \underbrace{\frac{n}{e^{n - e \log n}}}_{0}}$$
A: $$\lim_{n\to\infty} \sqrt[n]{n^e+e^n}=\lim_{n\to\infty} (n^e+e^n)^{\frac1n}=\lim_{n\to\infty} e^{\ln (n^e+e^n)\frac1n}$$
Now you just have to compute $$\lim_{n\to\infty} \frac{\ln (n^e+e^n)}n=\lim_{n\to\infty} \frac{\ln (e^{e\ln n}+e^n)}n=\lim_{n\to\infty} \frac{\ln e^n(e^{e\ln n-n}+1)}n=1+\lim_{n\to\infty} \frac{\ln (e^{e\ln n-n}+1)}n=1+\lim_{n\to\infty} \frac{\ln (e^{-\infty}+1)}n=1+\lim_{n\to\infty} \frac{\ln (0+1)}n=1+\lim_{n\to\infty} \frac{\ln (1)}n=1+0=1$$
since $e\ln n-n\to -\infty$.
A: Let's see a more direct way
$$\lim\limits_{n\to\infty} \sqrt[n]{n^e+e^n}=\lim\limits_{n\to\infty} \sqrt[n]{e^n}=e$$
because the ratio test applied to $\frac{n^e}{e^n}$ yields $\lim_{n\to\infty}\frac{n^e}{e^n}=0.$
A: You can also apply two gendarmes theorem. The idea is to find sequences $l_n, u_n$ which bound (from below and above) the given sequence $a_n$, i.e. $$l_n\le a_n\le u_n$$ holds, and which converge to a joint limit (i.e. $\lim_{n\to\infty}l_n=\lim_{n\to\infty}u_n=g).$ Then the theorem says that $a_n$ is also convergent and the limit is $g$. Let's see how it works.
Finding these bounds is usually pretty straightforward – lower bound is often obtained by simple missing some nonnegative terms. While seeking the upper bound, one have to remeber that the inequality have to be true only for all sufficiently large $n$'s. In this case one can write:
$$\sqrt[n]{0+e^n}\le\sqrt[n]{n^e+e^n}\le\sqrt[n]{e^n+e^n}$$ 
since $0\le n^e$ and $n^e\le e^n$ for sure if $n$ is sufficiently large. Next, we observe that 
$l_n:= \sqrt[n]{e^n}=e\longrightarrow e$ as well as
$u_n:=\sqrt[n]{2e^n}=e\sqrt[n]{2}\longrightarrow e\cdot 1 =e.$
The theorem yields the claim.
A: This is equivalent to showing that $\lim_{n \rightarrow \infty} \sqrt[n]{\frac {n^e}{e^n} + 1 } = 1$. 
This is clearly bounded below by 1. It is bounded above by $ 1 + \frac {n^e}{n e^n}$, which has a limit of 1 since polynomials grow much slower than exponentials.
A: You have $$\sqrt[n]{n^e+e^n}= \exp \left( \frac{\ln(n^e+e^n)}{n} \right)= \exp \left(1+ \frac{\ln \left( 1+ \frac{n^e}{e^n} \right)}{n} \right)$$ But $\ln \left( 1+ \frac{n^e}{e^n} \right) \sim \frac{n^e}{e^n}$ so you can conclude. 
A: Note that by L'Hospital$$\lim_{n\to\infty}\frac{\log (n^e+e^n)}{n}=\lim_{n\to\infty}\frac{\frac{1}{n^e+e^n}(n^{e-1}+e^n)}{1}=\lim_{n\to\infty}\frac{n^{e-1}}{n^e+e^n}+\lim_{n\to\infty}\frac{1}{1+\frac{n^e}{e^n}}$$ Now $n.n^{e-1}\leq n^e+e^n$, so $n^{e-1}/(n^e+e^n)\leq1/n $. Hence taking limit on both side, the first limit is zero. For the second one it can be proved that $n^{e+1}\leq e^n$ for all $n\geq n_0$, for some $n_0$. So $\lim_{n\to\infty}(n^e/e^n)\leq\lim_{n\to\infty}(1/n)=0$. Hence the second limit equals $1$. Now taking exponential the required limit evaluates to $e$.
A: Taking logs, you must show that
$$\lim_{n \rightarrow \infty} {\ln(n^e + e^n) \over n} = 1$$
Applying L'hopital's rule, this is equivalent to showing
$$\lim_{n \rightarrow \infty}{en^{e-1} + e^n \over n^e + e^n} = 1$$
Which is the same as
$$\lim_{n \rightarrow \infty}{e{n^{e-1}\over e^n} + 1 \over {n^e \over e^n}+ 1} = 1$$
By applying L'hopital's rule enough times, any limit of the form $\lim_{n \rightarrow \infty}{\displaystyle {n^a \over e^n}}$ is zero. So one has
$$\lim_{n \rightarrow \infty}{e{n^{e-1}\over e^n} + 1 \over {n^e \over e^n}+ 1} = {e*0 + 1 \over 0 + 1}$$
$$ = 1$$
(If you're wondering why you can just plug in zero here, the rigorous reason is that the function ${\displaystyle {ex + 1 \over y + 1}}$ is continuous at $(x,y) = (0,0)$.)
