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May 13th, 2019, 05:01 AM  #1 
Newbie Joined: Aug 2016 From: Romania Posts: 23 Thanks: 0  Proof for this power limit without differentiation https://ibb.co/WVC8BgJ I am stuck at the last step and I don't know how to go on.Can someone help me?What do I do from the last step? 
May 13th, 2019, 12:59 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,755 Thanks: 695 
In general $\sqrt{1w}\approx 1\frac{w}{2}$ for small $w$.

May 13th, 2019, 07:40 PM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 609 Thanks: 378 Math Focus: Dynamical systems, analytic function theory, numerics  
May 14th, 2019, 07:33 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,654 Thanks: 2632 Math Focus: Mainly analysis and algebra 
\begin{align*} \lim_{x \to \infty} \cos^{x^2} \frac1x &= \lim_{x \to \infty} \left(1  2\sin^2\frac1{2x} \right)^{x^2} & (\cos 2A &= 12\sin^2 A) \\ &= \lim_{x \to \infty} \left(1  2\sin^2\frac1{2x} \right)^{\left(\frac{1}{2\sin^2\frac1{2x}}\right) \left(2{x^2}{\sin^2\frac1{2x}}\right)} \\ &= \lim_{x \to \infty} \left(1  2\sin^2\frac1{2x} \right)^{ \frac12 \left( \frac{1}{2\sin^2\frac1{2x}} \right) \left( \frac{\sin\frac1{2x}}{\frac1{2x}} \right)^2} \\ &= \lim_{x \to \infty} \left(1  2\sin^2\frac1{2x} \right)^{ \left( \frac{1}{2\sin^2\frac1{2x}} \right) \frac12 \left( \frac{\sin\frac1{2x}}{\frac1{2x}} \right)^2} \\ &= \left( \frac1e \right)^{ \frac12 \cdot 1^2 } \\ &= e^{\frac12} \end{align*} Last edited by v8archie; May 14th, 2019 at 08:08 AM. 
May 14th, 2019, 01:22 PM  #5 
Newbie Joined: Aug 2016 From: Romania Posts: 23 Thanks: 0  Where does 12sin^2(1/x) come from when cos(1/x) is:,,sqrt(1sin^2(1/x)''? That part seems confusing. Can you explain how cos(1/x)=cos(2*1/x)?
Last edited by alex77; May 14th, 2019 at 01:24 PM. 
May 14th, 2019, 01:50 PM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,654 Thanks: 2632 Math Focus: Mainly analysis and algebra 
The identity used is given in the first line.


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differentiation, limit, power, proof 
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