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How find that limit $\lim_{x\to 0}\frac{\sin\left(x\sin\frac{1}x\right)}{x\sin\frac{1}x}$? Its $\lim_{x\to 0}\frac{\sin\left(x\sin\frac{1}x\right)}{x\sin\frac{1}x}=1$?

piteer
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3 Answers3

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Whatever $u$ tends to $0$ we have $$\lim_{u\to0}\frac{\sin u}{u}=1$$ In our case we take $u=x\sin\left(\frac1x\right)\xrightarrow{x\to0}0$ and the result follows.

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As $\displaystyle\left|\sin\frac1x\right|\le1, \lim_{x\to0}x\sin\frac1x=0$

lab bhattacharjee
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$$\lim_{x\to 0}\frac{\sin\left(x\sin\frac{1}x\right)}{x\sin\frac{1}x}=\frac{0}{0}$$

Using De L'Hospital's rule:

$$\lim_{x\to 0}\frac{\sin\left(x\sin\frac{1}x\right)}{x\sin\frac{1}x}=\lim_{x\to 0}\frac{\cos\left(x\sin\frac{1}x\right) \left(x\sin\frac{1}x\right) '}{\left(x\sin\frac{1}x\right ) '}=\\ \lim_{x\to 0}\cos\left(x\sin\frac{1}x\right) =\cos\left(\lim_{x\to 0} x\sin\frac{1}x\right)=(*)$$ $$$$ $$ |x\sin\frac{1}x | \leq |x| \Rightarrow -x \leq x \sin\frac{1}x \leq x$$

Using the Squeeze Thereom:

$$\lim_{x\to 0} x\sin\frac{1}x=0$$

Therefore, $$(*)=\cos(0)=1$$

Mary Star
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  • http://www.wolframalpha.com/input/?i=lim+sin+%28x+sin+1%2Fx%29+%2F+%28x+sin+1%2Fx%29+as+x+-%3E0 – piteer Aug 06 '14 at 12:13
  • What do you mean?? – Mary Star Aug 06 '14 at 12:14
  • in wolfram this limit is not 1 – piteer Aug 06 '14 at 12:22
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    @piteer Wolfram says the limit is undefined (which it would be, before you use L'Hospital's rule). Mary Star's method allows you to give a value to limits that are initially undefined. – Jam Aug 06 '14 at 12:24
  • i found http://math.stackexchange.com/questions/364552/strange-behavior-of-lim-x-to0-frac-sin-leftx-sin-left-frac1x-right-right?rq=1 – piteer Aug 06 '14 at 12:27