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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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    Ni$c$e to know that an experienced human can eyeball an expression, or a chess position, and a computer program can't (yet).2011-06-23

3 Answers 3

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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 q$u$estion.2011-06-23
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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.2011-06-23
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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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    You didn't waste my time. Its cool :) I don't like the use of l'Hospital every time, but when it's possible I use it, since it simplifies computations sometimes.2011-06-23