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

Please, help me in solving of $\lim\limits_{m\to\infty}\left(\cos\frac xm\right)^{m}$.

share|cite|improve this question
l'Hôpital's rule – Elements in Space Jan 23 '13 at 8:29
Just notice this $\cos(\frac{x}{m})\sim 1$ as $ m\to \infty $. – Mhenni Benghorbal Jan 26 '13 at 5:02
@MhenniBenghorbal Just notice this... Exactly the mistake to avoid. Note that $1\sim1$, $1+1/m\sim1$ and $1+1/\sqrt{m}\sim1$ while $1^m\to1$, $(1+1/m)^m\to\mathrm e$ and $(1+1/\sqrt{m})^m\to\infty$. – Did Jan 26 '13 at 5:09
@MhenniBenghorbal I thought about it, but $1^\infty$ is still indeterminate in that case. – Igor Gorbunov Jan 27 '13 at 12:35
up vote 3 down vote accepted

The approximation $\cos(x)\approx 1-\frac{x^2}{2}$ (which has several geometric proofs) and the binomial approximation $(1+t)^n\approx 1+nt$ are enough to give you $\displaystyle\left(\cos\frac{x}{m}\right)^m\approx\left(1-\frac{x^2}{2m^2}\right)^m\approx 1-\frac{x^2}{2m}$, and of course taking the limit as $m\to\infty$ in the latter expression is trivial.

share|cite|improve this answer
$1\approx 2$ but they don't have equal limits. – Michael Albanese Jan 23 '13 at 8:44
Thank you! This is fine. – Igor Gorbunov Jan 23 '13 at 9:02
Nice and fine answer +1 – Babak S. Jan 23 '13 at 9:05
@MichaelAlbanese That depends entirely on what your usage of $\approx$ is. I'm (informally and implicitly) using a version in which $a\approx b+c$ stands for $a=b+c+o(c)$; it's easy to formalize these manipulations and for a proof one would obviously want to be a lot more careful than I was here, but for a problem this simple I felt the extra notation just gets in the way. – Steven Stadnicki Jan 23 '13 at 16:36
Of course, I should have made myself a bit clearer. I wasn't doubting your method, I just thought the use of the symbol $\approx$ doesn't make it clear that the approximations need to be fairly good ones (in the sense that you have made clear in your comment). – Michael Albanese Jan 23 '13 at 20:17

Another way $$\lim\limits_{m\to\infty}\left(1+\left(\cos\frac xm-1\right)\right)^{m}=\lim\limits_{m\to\infty}e^{\frac{\left(\displaystyle\cos\frac xm-1\right)}{\displaystyle\left(\frac{x}{m}\right)^2}\times \displaystyle\frac{x^2}{m}}=\lim\limits_{m\to\infty}e^{\displaystyle-\frac{1}{2}\times\frac{x^2}{m}}=e^0=1$$


share|cite|improve this answer

The answer is 1. Try taylor expanding cos and using $\lim\limits_{x\to0} \frac{\log({1+x})}{x}=1$.

share|cite|improve this answer
Thank you! But is there a possibility to solve it without using l'Hôpital's rule or Taylor expanding, i.e. using trigonometric identities? – Igor Gorbunov Jan 23 '13 at 8:34
Well, instead of taylor expanding you can use $cos(2x)=1-2sin^2(x)$ and $ lim_{x\to 0} \frac{sin(x)}{x}=1$ – Ishan Banerjee Jan 23 '13 at 8:40

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.