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May 13th, 2019, 06:01 AM   #1
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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?
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May 13th, 2019, 01:59 PM   #2
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In general $\sqrt{1-w}\approx 1-\frac{w}{2}$ for small $w$.
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May 13th, 2019, 08:40 PM   #3
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Originally Posted by mathman View Post
In general $\sqrt{1-w}\approx 1-\frac{w}{2}$ for small $w$.
How does this avoid differentiation?
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May 14th, 2019, 08:33 AM   #4
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\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 &= 1-2\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*}
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Last edited by v8archie; May 14th, 2019 at 09:08 AM.
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May 14th, 2019, 02:22 PM   #5
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Quote:
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\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 &= 1-2\sin^2 A) \\
Where does 1-2sin^2(1/x) come from when cos(1/x) is:,,sqrt(1-sin^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 02:24 PM.
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May 14th, 2019, 02:50 PM   #6
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The identity used is given in the first line.
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