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$$ \lim _{x\to 0} \frac{x^2\sin(\frac{1}{x})}{\sin x}$$

Okay since sine is bounded ${x^2\sin(\frac{1}{x})} \ \ \to 0$

$\sin x\to 0$ Thus we can apply l'hospitals to it .

Applying l'Hospital's we get :

$$\lim_{x\to 0}\frac{\sin(\frac{1}{x})(1-2x)}{\cos x}$$ Here We can't find a way out? What would you advice me to do? Is there anyway to compute the limit?

Does this imply that the limit doesn't exist btw? No I don't think so, I think the conditions of the l'Hospital's are just not met.

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  • $\begingroup$ L'Hopital's rule $\endgroup$ – user378947 Dec 26 '16 at 21:31
  • $\begingroup$ @mathbeing No need for it here, though. $\endgroup$ – Clement C. Dec 26 '16 at 21:36
  • $\begingroup$ Small l, big H. I guess you need to be a frenchman to say why. $\endgroup$ – mathreadler Dec 26 '16 at 21:36
  • $\begingroup$ Is OP even allowed to use l'Hospital rule here? How would one solve the limit he is left over with? $\endgroup$ – Piotr Benedysiuk Dec 26 '16 at 21:46
  • $\begingroup$ @ClementC. Obviously no need for it (arguably never). It just bothered me to read 'l'hospital'. Now I am just confused. Is it correct to say 'l'Hospital'? $\endgroup$ – user378947 Dec 26 '16 at 21:47
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Easier is to break into a product of limits: $$ \lim_{x\to 0} \frac{x^2\sin\tfrac{1}{x}}{\sin x} = \lim_{x\to 0} \frac{x}{\sin x}\cdot \lim_{x\to0}\,x\sin\tfrac{1}{x} = 1\cdot 0 = 0. $$

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Hint: Can you first calculate the limit below?

$$\lim_{x\to0} \frac{x^2}{\sin\left(x\right)}$$

Afterwards, notice, as you yourself said, that the sine is bounded.

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1). Use the standard limit (derivative of $\sin$ at $0$) $$ \frac{\sin x}{x} \xrightarrow[x\to 0]{} 1 $$ so that $$ \frac{x}{\sin x} \xrightarrow[x\to 0]{} 1 $$ as well.

2). Rewrite $$ \frac{x^2 \sin\frac{1}{x}}{\sin x} = x\cdot \frac{x}{\sin x}\cdot \sin\frac{1}{x} $$

3). Look at the three factors. The one goes to $0$, the second converges to $1$, the last is bounded.

4). Conclude.

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Recall that

$$\lim_{x\to0}\frac{\sin(x)}x=1$$

and

$$-x<x\sin(1/x)<x$$

Thus,

$$\lim_{x\to0}\frac{x^2\sin(1/x)}{\sin(x)}=\lim_{x\to0}\frac x{\sin(x)}x\sin(1/x)=0$$

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You might know that $\lim_{x\to 0} \frac{x}{\sin(x)} = 1$ and that $\lim_{x\to 0} x \sin(1/x) = 0.$

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L Hopital rule says that this is $$ \frac{(x^2\sin(\frac{1}{x}))'}{\cos x}$$ in cero, but the numerator has a discontinuity in the derivate, and the value must be calculated by definition: $$\lim _{x\to 0} \frac{x^2\sin(\frac{1}{x})}{x}=\lim _{x\to 0} x\sin(\frac{1}{x})=0$$ So the limit is cero. Another way: $$\lim _{x\to 0} \frac{x^2\sin(\frac{1}{x})}{\sin x}=\lim _{x\to 0}\frac{x}{sin(x)} x\sin(\frac{1}{x})=1*\lim _{x\to 0} x\sin(\frac{1}{x})=0$$

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We only need to know that $\lim_{x\to0}\frac{\sin x}{x}=1$, therefore $$\lim_{x\to0}\frac{x^2\sin\frac1x}{\sin x}=\lim_{x\to 0}\frac{x}{\sin x}x\sin \frac1x\sin x=\lim_{x\to 0}\underbrace{\frac{x}{\sin x}}_{\to 1}\underbrace{\frac{\sin \frac1x}{\frac1x}}_{\to 1}\underbrace{\sin}_{\to 0} x=0.$$

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