# Limit proof

1. Mar 28, 2005

If we are asked to prove the following: $$\lim_{x\rightarrow 0} \frac{x^{2}\sin\frac{1}{x}}{\sin x} = 0$$ would I do: $$\lim_{x\rightarrow 0} x^{2}(\sin\frac{1}{x})(\frac{1}{\sin x})$$? So the leading factor (dampening factor) approaches 0 which makes the whole expression approach 0?

Thanks

2. Mar 28, 2005

### robphy

It's probably better to use L'Hopital's Rule.

Note that you see some expressions of the form (sin y)/y, which may help you.

3. Mar 28, 2005

### dextercioby

Use the limits

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

$$\lim_{x\rightarrow +\infty} \frac{\sin x}{x}=0$$

Daniel.

4. Mar 28, 2005

So you have: $$\lim_{x\rightarrow 0} x^{2}(\frac{\sin x}{x})(\frac{1}{\sin x})$$ . And you get $$\frac{x^{2}}{\sin x}$$. So the limit is then 0?

Last edited: Mar 28, 2005
5. Mar 28, 2005

### robphy

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