Can the Limit of a Product of Fractions Approach 0?

[courtrigrad]

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

 

[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.

 

[dextercioby]

Use the limits

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

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

Daniel.

 

[courtrigrad]

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?

 

[robphy]

Check your algebra:

\lim_{x\rightarrow 0} \frac{x^{2}\sin(\frac{1}{x})}{\sin x} =<br /> \lim_{x\rightarrow 0} \frac{x}{\sin x} \frac{\sin(\frac{1}{x})}{\frac{1}{x}} <br />

 

[dextercioby]

The limit is 0,but the "sines" cannot be simplified.It's a product of 2 fractions,one o them goes to 1 & the other to 0.

Daniel.