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How should one go about computing $$\lim_{n\to\infty}{\left(1+\frac{2n^2+\cos{n}}{n^3+n}\right)^n}\quad?$$ What surprised me about this is that $$\lim_{n\to\infty}{\left(1+\frac{2n^2+\cos{n}}{n^3+n}\right)^\frac{n^3+n}{2n^2+\cos{n}}}=1$$(according to wolfram), instead of $e$, which is what I expected. Could someone comment on that also?

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Wolfram says $1$ and Maple says $e$. What do YOU say? –  GEdgar Jun 23 '11 at 19:37
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The standard (though sometimes labour-intensive) method of solving these indeterminate limits is to transform $f(x)^{g(x)}$ into $\exp(g(x)\ln(f(x)))$, and then compute the limit of $g(x)\ln(f(x))$. –  Arturo Magidin Jun 23 '11 at 19:44
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Nice to know that an experienced human can eyeball an expression, or a chess position, and a computer program can't (yet). –  André Nicolas Jun 23 '11 at 19:50

3 Answers 3

up vote 5 down vote accepted

Hint: Show that $$ \lim _{n \to \infty } n\ln \bigg(1 + \frac{{2n^2 + \cos n}}{{n^3 + n}}\bigg) = 2. $$

Elaborating: $$ n\ln \bigg(1 + \frac{{2n^2 + \cos n}}{{n^3 + n}}\bigg) = n\ln \bigg(1 + \frac{{2 + \cos (n)/n^2 }}{{n + 1/n}}\bigg) = na_n \frac{{\ln (1 + a_n )}}{{a_n }}, $$ where $$ a_n = \frac{{2 + \cos (n)/n{}^2}}{{n + 1/n}}. $$ Noting that $a_n \to 0$ as $n \to \infty$ and $$ \mathop {\lim }\limits_{x \to 0} \frac{{\ln (1 + x)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{1/(1 + x)}}{1} = 1, $$ we conclude that $$ \lim _{n \to \infty } \frac{{\ln (1 + a_n )}}{{a_n }} = 1 $$ and, in turn, $$ \lim _{n \to \infty } n\ln \bigg(1 + \frac{{2n^2 + \cos n}}{{n^3 + n}}\bigg) = \lim _{n \to \infty } na_n = \lim _{n \to \infty } n\frac{{2 + \cos (n)/n^2 }}{{n + 1/n}} = 2. $$ Thus, $$ \mathop {\lim }\limits_{n \to \infty } \bigg(1 + \frac{{2n^2 + \cos n}}{{n^3 + n}}\bigg)^n = e^2 . $$

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This refers to the first question. –  Shai Covo Jun 23 '11 at 19:44

Lets consider the more general case of an arbitrary function. Then we have the following theorem:

Theorem: Given a function $g(n)$ such that $g_0=\lim_{n\rightarrow \infty} g(n)$ exists, we have $$\lim_{n\rightarrow \infty }\left(1+\frac{g(n)}{n}\right)^n=e^{g_0}.$$

Example: In particular, for your above question, $g(n)=\frac{2n^2+\cos(n)}{n^2+1}$, so that $g_0 =2$, and hence the value of the original limit is $e^2$.

Proof of theorem: Let $f(n)=\frac{g(n)}{g_0}$ so that $f(n)=1+o(1)$. Then since $\lim_{n\rightarrow \infty} \left(1+\frac{a}{n}\right)^n=e^a$ we see that $$\lim_{n\rightarrow \infty} \left(1+\frac{g(n)}{n}\right)^\frac{n}{f(n)}=e^{g_0}.$$ Then, because $f(n)=1+o(1)$, it follows that $\frac{n}{f(n)}=n+o(n)$. But, $$\lim_{n\rightarrow \infty} \left(1+\frac{O(1)}{n}\right)^{o(n)}=1,$$ so we conclude that $$\lim_{n\rightarrow \infty }\left(1+\frac{g(n)}{n}\right)^n=e^{g_0}.$$

I hope that helps,

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This is a nice example of a problem in which the generalization is easier to prove than the case in question. –  Mark Jun 23 '11 at 21:13

What you should use is that $$ \lim_{x \to 0} (1+x)^{\frac{1}{x}}=e$$

Then, if you have to calculate $\lim_{n \to \infty} x_n^{y_n}$ where $x_n \to 1$ and $y_n \to \infty$ you proceed as follows:

  • Denote $a_n=x_n-1$ so $a_n \to 0$.

  • Now you have to calculate $\lim_{n \to \infty} (1+a_n)^{y_n}$.

  • Transform the exponent so that you get the $e$-limit presented above: $$ \lim_{n \to \infty} ((1+a_n)^{\frac{1}{a_n}})^{a_n y_n} =e^L$$ where $\displaystyle L=\lim_{n \to \infty}a_n y_n$, which usually is pretty simple to calculate.

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$\displaystyle\lim_{n->\infty}a_ny_n=\displaystyle\lim_{n->\infty}\frac{2n^2+\co‌​s{n}}{n^2+1}=2$ hence the final answer $e^2$ in this case? Is there an argument as to why we are allowed to consider the $\displaystyle\lim_{n->\infty}a_ny_n$ separately? –  Julius Jun 23 '11 at 20:04
    
$a_n \to 0$ need not imply $(1+a_n)^{1/a_n} \to e$. In this case ($\frac{2n^2 + \cos n}{n^3 + n}$), it works, as the terms are eventually positive. –  Aryabhata Jun 23 '11 at 20:11
    
@Aryabahata: Why it does not imply that fact? $\lim_{x \to 0}\frac{\ln(x+1)}{x}=1$ by l'Hospital's rule, whenever $x$ is negative or positive, as long as $1+x$ is positive, and it is eventually, since $x \to 0$. –  Beni Bogosel Jun 23 '11 at 20:21
    
@Beni: I think I am having one of those "duh" moments :-). I was thinking of $\lim 1/a_n$ rapidly oscillating between $-\infty$ and $+\infty$, but you are probably right. –  Aryabhata Jun 23 '11 at 20:23
    
You are right as $f(x) = (1+x)^{1/x}$ with $f(0) = e$ is continuous at $0$. I really don't like using L'Hospital rule for some reason. (btw, you already have my upvote). Apologies for wasting your time. –  Aryabhata Jun 23 '11 at 20:34

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