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 August 21st, 2010, 01:15 AM #1 Senior Member   Joined: Aug 2010 Posts: 109 Thanks: 0 find the limit
 August 21st, 2010, 02:36 AM #2 Member   Joined: Jul 2010 Posts: 95 Thanks: 0 Re: find the limit I would do it like that $\lim_{x\to\0}( \cos x)^{\frac{1}{x^2}}$ would be something like $\lim_{x\to\0}( \frac{\cos x}{x^2})$ then $\frac{\lim_{x\to\0} \cos x}{\lim_{x\to\0} x^2$ and then $\frac{\lim_{x\to\0} \cos x}{(\lim_{x\to\0} x)^2}=\infty$ but I guess I am completely wrong
 August 21st, 2010, 01:13 PM #3 Member   Joined: Aug 2010 Posts: 49 Thanks: 0 Re: find the limit you are completely wrong! Since $(\cos\frac{1}{n})^{n^2}<(\cos\frac{1}{n})^n<1$ (because the cosine function is less than 1 for every $n\in\mathbb{N}$) the limit is between 0 and 1. Your application of rules for limits (and mathematics) is wrong - but nevertheless, I am thinking about the right solution. It will be 1... and I hope someone is faster, since I forgot a lot since studying physics. Regards, Jens
 August 21st, 2010, 01:13 PM #4 Global Moderator   Joined: May 2007 Posts: 6,307 Thanks: 526 Re: find the limit Let T = (cosx)^1/x^2 = exp[ln(cosx)/x^2]. As x ->0, cosx ~ 1 - x^2/2, so ln(cosx) ~ - x^2/2, therefore T ~ exp(-1/2).
 August 21st, 2010, 04:44 PM #5 Newbie   Joined: Aug 2010 Posts: 17 Thanks: 0 Re: find the limit It is my first time, and I worked on the answer, and It is: 1/Sqrt[e]. I hope it is the Right answer. Let me know if I am wrong.
 August 21st, 2010, 05:58 PM #6 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,575 Thanks: 931 Math Focus: Elementary mathematics and beyond Re: find the limit $\lim_{x \to 0}\,(\cos{x})^{1/x^2}$ $=\lim_{x \to 0}\,e^{\ln((\cos{x})^{1/x^2})$ $=\lim_{x \to 0}\,e^{\frac{\ln(\cos{x})}{x^2}$ l'Hopital's rule: $=\lim_{x \to 0}\,e^{\frac{-\sin{x}}{2x\cos{x}}$ and again l'Hopital's: $=\lim_{x \to 0}\,e^{\frac{-cos{x}}{2\cos{x}\,-\,2x\sin{x}}}=\frac{1}{\sqrt{e}}$
 August 21st, 2010, 06:00 PM #7 Member   Joined: Aug 2010 Posts: 49 Thanks: 0 Re: find the limit right! Not a prove but indeed the right way to prove it, I guess. Sorry for the 1 posted.
 August 21st, 2010, 06:08 PM #8 Member   Joined: Aug 2010 Posts: 49 Thanks: 0 Re: find the limit my posting was with respect to mathman, l'Hospital is a perfect hit.

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