Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

What would be the limit of $\cos(x/n)^{n^2}$ as $n\to\infty$ and how would you proceed with the proof?

I tried to use the exponential form of cosine: $$\frac{[e^{ix/n} + e^{-ix/n}]^{n^2}}{2^{n^2}}.$$

However, what is bothering me is that how I should get rid of the $n^2$ in the power. Please help. Thanks.

share|cite|improve this question
This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Did Feb 20 '14 at 7:50
I would use the fact that for large $n$ we have $\cos(x/n)\approx 1-\frac{x^2}{2}\cdot\frac{1}{n^2}$. – André Nicolas Feb 20 '14 at 7:57
I figured that I could may be write cos(x/n) in terms of exponential function such as [e^(ix/n) + e^(-ix/n)]/2, but I am not able to proceed any further as I don't know how to manipulate the power n^2 to get to an answer. – user130113 Feb 20 '14 at 7:58
Also x is any real number. Sorry forgot to mention that! – user130113 Feb 20 '14 at 7:59
Use my Taylor series suggestion. The limit is $e^{-x^2/2}$. You can also alternately express the function as $\exp(n^2\ln(\cos(x/n)))$ and use L'Hospital's Rule. – André Nicolas Feb 20 '14 at 8:00
up vote 1 down vote accepted

Hint : Take the logarithm, and apply l'Hopital.

share|cite|improve this answer
yeah I figured this one after all with all the help provided in the comments box. Thanks so much anyway. – user130113 Feb 20 '14 at 22:14

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.