Question

find the limit as x approaches 0 of (sin^2)(4x)/3x

Answer 1

Do you know L'Hopital's Rule?

Answer 2

@PlasmaFuzer no

Answer 3

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

Answer 4

Oh yes i know that

Answer 5

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

Answer 6

Ahh yes squeeze theorem good one @jim_thompson5910

Answer 7

\[\sin^2(u)=\sin(u) \cdot \sin(u)\]

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

@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?

Answer 12

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

Answer 13

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

Answer 14

correct

Answer 15

Great! so the answer is 0. thanks!

Answer 16

you're welcome