Calculating the limit of $\frac{x^2\sin(\frac{1}{x})}{\sin x}$ $$ \lim _{x\to 0} \frac{x^2\sin(\frac{1}{x})}{\sin x}$$
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.
 A: Easier is to break into a product of limits:
$$
\lim_{x\to 0} \frac{x^2\sin\tfrac{1}{x}}{\sin x} = \lim_{x\to 0} \frac{x}{\sin x}\cdot \lim_{x\to0}\,x\sin\tfrac{1}{x} = 1\cdot 0 = 0.
$$
A: 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.
A: 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.
A: Recall that
$$\lim_{x\to0}\frac{\sin(x)}x=1$$
and
$$-x<x\sin(1/x)<x$$
Thus,
$$\lim_{x\to0}\frac{x^2\sin(1/x)}{\sin(x)}=\lim_{x\to0}\frac x{\sin(x)}x\sin(1/x)=0$$
A: You might know that $\lim_{x\to 0} \frac{x}{\sin(x)} = 1$ and that $\lim_{x\to 0} x \sin(1/x) = 0.$ 
A: 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:
$$\lim _{x\to 0} \frac{x^2\sin(\frac{1}{x})}{\sin x}=\lim _{x\to 0}\frac{x}{sin(x)} x\sin(\frac{1}{x})=1*\lim _{x\to 0} x\sin(\frac{1}{x})=0$$
A: We only need to know that $\lim_{x\to0}\frac{\sin x}{x}=1$, therefore 
$$\lim_{x\to0}\frac{x^2\sin\frac1x}{\sin x}=\lim_{x\to 0}\frac{x}{\sin x}x\sin \frac1x\sin x=\lim_{x\to 0}\underbrace{\frac{x}{\sin x}}_{\to 1}\underbrace{\frac{\sin \frac1x}{\frac1x}}_{\to 1}\underbrace{\sin}_{\to 0} x=0.$$
