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Solve $\lim_{x \to 0}x\sin(\frac{1}{x})$ Using $\lim_{\theta \to 0}\frac{\sin \theta}{\theta} = 1$

I have solved it using the Squeeze Theorem and now I want to solve it using $\lim_{\theta \to 0}\frac{\sin \theta}{\theta} = 1$.

Why can't I multiply the numerator and denominator by $\frac{1}{x}$ so I can have it in the form of $\lim_{x \to 0}\frac{\frac{1}{x}x\sin(\frac{1}{x})}{\frac{1}{x}} = \lim_{x \to 0}\frac{\sin(\frac{1}{x})}{\frac{1}{x}}$ so that I can apply $\lim_{\theta \to 0}\frac{\sin \theta}{\theta} = 1$.

I understand that $\lim_{x \to 0}\frac{1}{x}$ does not exist, so does that mean that I can't solve the above mentioned using $\lim_{\theta \to 0}\frac{\sin \theta}{\theta} = 1$?

Kermit the Hermit
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  • The fact that $\lim_{\theta\to 0} \frac{\sin\theta}{\theta}=1$ is not relevant to the problem. Squeezing is the way to go. – André Nicolas Aug 08 '14 at 15:09
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    I was just curious to see if I can apply that type of method after I already solved it using the Squeeze Theorem. So would you say that there's no way for me to solve it using $\lim_{\theta \to 0}\frac{\sin(\theta)}{\theta}$? – Kermit the Hermit Aug 08 '14 at 15:15
  • I understand now, thanks, @AndréNicolas. – Kermit the Hermit Aug 08 '14 at 15:22
  • Yes, I would say there is no way. We need the behaviour of $\sin z$ for large $z$, and the behaviour of $\sin t$ for small $t$ is not relevant. Note for example that $\lim_{t\to 0}\frac{e^t-1}{t}=1$, but the behaviour of $x(e^{1/x}-1)$ is entirely different from the behaviour of $x\sin(1/x)$ for $x$ near $0$. – André Nicolas Aug 08 '14 at 15:23
  • The limit is $0$ as $sin\frac1x$ is bounded. –  Aug 08 '14 at 16:04
  • Related (proof using squeeze theorem|: http://math.stackexchange.com/questions/1066434/calculate-lim-x-to-0-x-cdot-sin-frac1x – Martin Sleziak Nov 19 '15 at 14:35

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The problem is that you would need to have $\frac1x\rightarrow 0$, but that's not the case.

And it's even worse:

  • as $x\rightarrow 0+$, you have $\frac1x\rightarrow +\infty$

  • as $x\rightarrow 0-$, you have $\frac1x\rightarrow -\infty$

MPW
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  • Yes, that much I understand hence including "I understand that $\lim_{x \to 0}\frac{1}{x}$ does not exist, so does that mean that I can't solve the above mentioned using $\lim_{\theta \to 0}\frac{\sin(\theta)}{\theta}$?", or am I misunderstanding what you're saying? – Kermit the Hermit Aug 08 '14 at 15:10
  • What I mean is you don't have $\theta\rightarrow 0$, so that formula doesn't apply. If the problem were something like $$\lim_{x\rightarrow\infty}x\sin\frac1x$$ you could used that formula. But it isn't. You seem to be overlooking that the formula requires the expression in the denominator, which is also the argument of the sine, must approach zero to apply. – MPW Aug 08 '14 at 15:14
  • Now I get it, indeed that is what I was overlooking. Thanks for the insight, @MPW, I appreciate it. – Kermit the Hermit Aug 08 '14 at 15:21
  • You're quite welcome. Glad to be of assistance. :) – MPW Aug 09 '14 at 07:40