# Converting limit to e exponent form

I don't understand why this is true. I would like to see the process of converting this limit to e, to better understand this topic.

$$\lim_{n \to \infty }\left({n^{2}+n \over n^{2}+1}\right)^{n} = e^1 = e$$

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I have inserted some brackets for clarity, remove them if they are wrong – Henry Nov 27 '11 at 0:07

Note that $$({n^{2}+n \over n^{2}+1})^{n}=({1+\frac{1}{n} \over 1+\frac{1}{n^{2}}})^n=\frac{(1+\frac{1}{n})^n}{(1+\frac{1}{n^{2}})^n}.$$ Note also that $$\lim_{n \to \infty }(1+\frac{1}{n})^n=e.$$ On the other hand, $$\lim_{n \to \infty }(1+\frac{1}{n^{2}})^n=\lim_{n \to \infty }e^{n\ln(1+\frac{1}{n^{2}})} =e^{\lim_{n \to \infty }\frac{\ln(1+\frac{1}{n^{2}})}{\frac{1}{n}}}.$$ By L'Hosiptal Rule, $$\lim_{n \to \infty }\frac{\ln(1+\frac{1}{n^{2}})}{\frac{1}{n}}=\lim_{n \to \infty }\frac{(\frac{1}{1+\frac{1}{n^{2}}})(-\frac{2}{n^3})}{(-\frac{1}{n^2})}=0.$$ Combining the above the equalities, we have $$\lim_{n \to \infty }(1+\frac{1}{n^{2}})^n=e^0=1.$$ Therefore, $$\lim_{n \to \infty }({n^{2}+n \over n^{2}+1})^{n} = \frac{\lim_{n \to \infty }(1+\frac{1}{n})^n}{\lim_{n \to \infty }(1+\frac{1}{n^{2}})^n}=\frac{e}{1}=e.$$

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Thank you, I now understand. I have to say, this looks so elegant. – Dave Nov 27 '11 at 13:39
Good that it is helpful to you. – Paul Nov 27 '11 at 13:57
@Matt, Paul is using l'Hopital - you told Jeroen you don't know l'Hopital, so what's up? Paul, see my comment on Jeroen's answer, regarding the possible circularity in the use of l'Hopital here. – Gerry Myerson Nov 27 '11 at 22:31

Are you familiar with $$\lim_{n\to\infty}\left(1+{1\over n}\right)^n=e$$ If so, do you see how to compare the limit you want with the one you know?

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Let $y=\lim_{x\to\infty}(1+\frac{1}{x})^x$ then:

\begin{align*}\ln y &= \ln \space[\lim_{x \to \infty} (1+\frac{1}{x})^x] \\ &= \lim_{x \to \infty}[\ln (1+\frac{1}{x})^x] \\ &= \lim_{x \to \infty}[x\ln (1+\frac{1}{x})] \\ &= \lim_{x \to \infty}[ \frac{\ln (1+\frac{1}{x})}{\frac{1}{x}}]\end{align*}

L'Hopital's rule is now used because other wise the limit would result in $\frac{0}{0}$:

\begin{align*} \ln y &= \lim_{x \to \infty}[\frac{(\frac{\frac{-1}{x^2}}{1+\frac{1}{x}})}{\frac{-1}{x^2}}] \\ &= \lim_{x \to \infty} 1+\frac{1}{x} \\ &= 1\end{align*}

We now have $\ln y = 1$, this implies $y=e$. This concludes the proof.

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This may be circular in that the formula for the derivative of the logarithm function may rest on the very limit you are trying to evaluate. – Gerry Myerson Nov 27 '11 at 0:07
Whoops I was going out from the title too much when giving this answer. Hope it still serves some purpose. EDIT: @GerryMyerson looks like you got me. I always thought it was a cool way of tackling it, but you are right about the circularity. – user12205 Nov 27 '11 at 0:16
I don't know L'Hôpital's rule, yet. This is a little bit hard for me. – Dave Nov 27 '11 at 0:28

Another way to show that $\lim_{n \to \infty }(1+\frac{1}{n^{2}})^n=1$ is to use this "contra-Bernoulli-inequality" (as I call it):

If $n$ is a positive integer and $0 \le x < 1/n$ then $(1+x)^n \le 1/(1-x n)$.

This is readily proved by induction, writing it in the form $(1-x n)(1+x)^n \le 1$.

Setting $x = 1/n^2$, we get $$(1+1/n^2)^n \le 1/(1-1/n) = 1+\frac{1}{n-1}.$$

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