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We could use L'Hospital here, because both numerator as well as denominator tend towards 0, I guess. The derivative of the numerator is $$x^2\cdot \left(-\sin\left(\frac{1}{x}\right)\right) \cdot \left( -\frac{1}{x^2}\right) + 2x \cos\left(\frac{1}{x}\right)=\sin\left(\frac{1}{x}\right) + 2x \cos\left(\frac{1}{x}\right) $$ The derivative of the denominator is $\cos(x)$. So, $$\lim\limits_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)} = \lim\limits_{x\rightarrow 0}\displaystyle\frac{\sin\left(\frac{1}{x}\right) + 2x \cos\left(\frac{1}{x}\right)}{\cos(x)}$$

Is that right so far?

Thanks for the help in advance. Best Regards, Ahmed Hossam

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  • $\begingroup$ The numerator is essentially $\pm x^2$ and the numerator $x$. $\endgroup$ – Yves Daoust Dec 16 '18 at 20:57
  • $\begingroup$ For all practical purposes this is a duplicate of 1, 2 and 3 proving once again that calculus answering machines think that the site rules don't apply to them. $\endgroup$ – Jyrki Lahtonen Dec 17 '18 at 7:15
  • $\begingroup$ Well, $\lim\limits_{x\to0}{\frac{x^2\cdot\sin\frac{1}{x}}{\sin x}}$ ist not the same as $\lim\limits_{x\to0}{\frac{x^2\cdot\cos\frac{1}{x}}{\sin x}}$ $\endgroup$ – Ahmed Hossam Dec 20 '18 at 10:04
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As we know that $$\lim_{x\to 0}{x\over \sin x}=1$$therefore $$\lim_{x\to 0}{x^2\cos {1\over x}\over \sin x}=\lim_{x\to 0}x\cos{1\over x}=\lim_{u\to \infty}{\cos u\over u}=0$$

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We don’t need l’Hopital, indeed by standard limits

$$\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)}=\frac{x \cos\left(\frac{1}{x}\right)}{\frac{\sin(x)}x}\to \frac01=0$$

indeed by squeeze theorem

$$\left|x \cos\left(\frac{1}{x}\right)\right|\le |x|\to0$$

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  • $\begingroup$ Thanks for your answer. $\endgroup$ – Ahmed Hossam Dec 20 '18 at 10:05
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Hint: Without using L’Hôpital’s Rule, note that

$$\frac{x^2\cos\big(\frac{1}{x}\big)}{\sin x} = \frac{x}{\sin x}\cdot x\cos\bigg(\frac{1}{x}\bigg)$$

and recall $\lim_\limits{x \to 0}\frac{\sin x}{x} = 1$.

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  • $\begingroup$ order of zero is higher in Numerator then in denominator hemce 0 $\endgroup$ – maveric Dec 16 '18 at 18:11
  • $\begingroup$ Thanks for your answer. $\endgroup$ – Ahmed Hossam Dec 20 '18 at 10:06
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With the Taylor power series, $\sin x= x+o(x)$ $$\lim_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)}=\lim_{x\to0}{x \cos\left(\frac{1}{x}\right)}=0$$ Because $x\to0$ and $\cos(1/x)$ is bounded from $-1$ to $1$ as $x\to0$

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  • $\begingroup$ Thanks for your answer. $\endgroup$ – Ahmed Hossam Dec 20 '18 at 10:06

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