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Why is $\lim\limits_{n\to\infty} \sqrt[n]{n^e+e^n}$ = $e$? I couldn't get this result.

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  • $\begingroup$ I tried it with the triangle inequality and did not get the right result. I tried it with $a^b = e^{a*ln(b)}$ but did not get any further. This is a exercise I do for exam preparation, it's not homework. $\endgroup$ – leo Dec 29 '12 at 11:06
  • $\begingroup$ Thanks for all the brilliant answers. The accepted one is the one which I could understand the best, even if might not be the most elegant or shortest. $\endgroup$ – leo Dec 29 '12 at 13:57
  • $\begingroup$ Would you like me to provide more details about my solution? What do you not understand about it? It's a straightforward way to deal with n roots, and is just based on Bernoulli's inequality that $(1+x)^n \geq 1+ nx$ for $x\geq 0, n\in \mathbb{N}$. $\endgroup$ – Calvin Lin Dec 30 '12 at 1:08
  • $\begingroup$ @CalvinLin I changed my mind now. Nameless solution did just use Math I was more familiar with. But your solution is very clean and I understand it now. $\endgroup$ – leo Jan 27 '13 at 16:31
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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.

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$$ \large (n^e + e^n)^{\frac{1}{n}} = e \left ( 1 + \frac {n^e}{e^n} \right) ^{\frac 1 n} $$ $$ e \Large \left ( 1 + \frac {n^e}{e^n} \right) ^{\frac 1 n} = e \left ( \underbrace { \left ( 1 + \frac {1}{e^{n - e \log n}} \right) ^{e^{n - e \log n}} }_{e}\right)^{ \underbrace{\frac{n}{e^{n - e \log n}}}_{0}}$$

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    $\begingroup$ What BIG fonts you have!! $\endgroup$ – Haskell Curry Dec 30 '12 at 8:13
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$$\lim_{n\to\infty} \sqrt[n]{n^e+e^n}=\lim_{n\to\infty} (n^e+e^n)^{\frac1n}=\lim_{n\to\infty} e^{\ln (n^e+e^n)\frac1n}$$ 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 $e\ln n-n\to -\infty$.

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  • $\begingroup$ could you please show the second last step a bit more fine grained. Otherwise this is the most clear answer to me. $\endgroup$ – leo Dec 29 '12 at 12:56
  • $\begingroup$ @leo Sure I can $\endgroup$ – Nameless Dec 29 '12 at 12:57
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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 $\lim_{n\to\infty}\frac{n^e}{e^n}=0.$

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  • $\begingroup$ Why is $n^e$ is negligible? $\endgroup$ – leo Dec 29 '12 at 11:28
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    $\begingroup$ @leo: because the exponential function grows much faster than the polynomial function. $\endgroup$ – user 1357113 Dec 29 '12 at 11:29
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    $\begingroup$ That is of course far from rigorous. $\endgroup$ – Eckhard Dec 29 '12 at 12:19
  • $\begingroup$ @Eckhard: in addition, this is the way I was taught to answer such questions in high school. $\endgroup$ – user 1357113 Dec 29 '12 at 12:30
  • $\begingroup$ I have to agree this is far from rigorous. $\endgroup$ – user641 Dec 29 '12 at 12:54
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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.

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  • $\begingroup$ You can also use that $\sqrt[n]{a+b}\le \sqrt[n]{a}+\sqrt[n]{b}$. $\endgroup$ – user641 Dec 29 '12 at 19:31
  • $\begingroup$ Another very nice answer! Thank you! $\endgroup$ – leo Dec 29 '12 at 20:07
  • $\begingroup$ $0\le n^e$ and $n^e\le e^n$ are true for all $n\ge 0$ $\endgroup$ – Henry Dec 29 '12 at 22:44
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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.

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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 $e$.

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

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