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.

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 $(x,y) = (0,0)$.)
