Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Wolfram says $1$ and Maple says $e$. What do YOU say? – GEdgar Jun 23 '11 at 19:37
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
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
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, $$ \mathop {\lim }\limits_{n \to \infty } \bigg(1 + \frac{{2n^2 + \cos n}}{{n^3 + n}}\bigg)^n = e^2 . $$

share|cite|improve this answer
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,

share|cite|improve this answer
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.

share|cite|improve this answer
$\displaystyle\lim_{n->\infty}a_ny_n=\displaystyle\lim_{n->\infty}\frac{2n^2+\co‌​s{n}}{n^2+1}=2$ 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.