Why does using the substitution u=x^2 help evaluate the limit as x approaches 0?

[JustinLiang]

Homework Statement

http://www.prep101.com/files/Math100PracticeExam.pdf

Question 1m

The Attempt at a Solution

I tried to do this by using
lim sinx/x = 1
x->0

Factoring out x^2 from each one and I get infinity.
Why does this method not work?

The answer says it's 4/5.

I even tried plugging in 0.001 in my calculator and I get infinity :S.

 

[QuarkCharmer]

Is that sin(2x^2), sin^2(2x), sin(2)x^2 ?

 

[Dick]

Is that sin(2x^2), sin^2(2x), sin(2)x^2 ?

Or is it 3sin(2x^2-6x^2+4x^6)? The question isn't very grammatical. And however I rearrange it I still can't figure out how to pull 4/5 as a limit out.

 

[JustinLiang]

Or is it 3sin(2x^2-6x^2+4x^6)? The question isn't very grammatical. And however I rearrange it I still can't figure out how to pull 4/5 as a limit out.

It is 3sin(2x^2), you get 4/5, this limit is INSANE :P

 

[vela]

It helps to use the substitution u=x² to get
$$\lim_{u \to 0}\frac{3 \sin 2u - 6u + 4u^3}{u^5}$$Then three applications of the Hospital rule gets you to a limit you can evaluate.

 

[Dick]

It helps to use the substitution u=x² to get
$$\lim_{u \to 0}\frac{3 \sin 2u - 6u + 4u^3}{u^5}$$Then three applications of the Hospital rule gets you to a limit you can evaluate.

Ah, indeed it does.