## anonymous one year ago find the limit as x approaches 0 of (sin^2)(4x)/3x

1. anonymous

Do you know L'Hopital's Rule?

2. anonymous

@PlasmaFuzer no

3. jim_thompson5910

Hint: $\Large \lim_{x \to 0}\left(\frac{\sin(x)}{x}\right) = 1$

4. anonymous

Oh yes i know that

5. anonymous

I'm just not sure how to simplify it to make it look like that

6. anonymous

Ahh yes squeeze theorem good one @jim_thompson5910

7. freckles

$\sin^2(u)=\sin(u) \cdot \sin(u)$

8. anonymous

I've gotten so used to L'Hopital's rule I'm lazy

9. jim_thompson5910

I suggest first breaking up the square as freckles wrote $\Large \frac{\sin^2(4x)}{3x}=\frac{\sin(4x)}{3}*\frac{\sin(4x)}{x}$ then try to make the denominator x turn into 4x

10. anonymous

Btw once you have completed the problem using jim's method I would suggest you check this out: https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule as you will eventually use it. It makes problems like this just a simple matter of taking derivatives. Though if you haven't seen it in class I would recommend against using it in the solution for this problem in particular, although if you fell able you should try it out and compare the effort it takes using both methods :D

11. anonymous

@jim_thompson5910 so I am able to get to $\frac{ 4 }{ 3 }\lim_{x \rightarrow 0}\frac{ \sin x }{ 1 } \times \frac{ \sin x }{ x }$ but how do I get an x into the first denominator?

12. jim_thompson5910

there's no need because you can simply plug in x = 0 for that first fraction

13. anonymous

so then would the answer be $\frac{ 4 }{ 3 }\times0\times1$

14. jim_thompson5910

correct

15. anonymous

Great! so the answer is 0. thanks!

16. jim_thompson5910

you're welcome