# Homework Help: Find out the limit of the following function

1. Apr 12, 2012

### vkash

f(0,∞)->R
f(x)=2 [x^(sin(2x)] cos(2x)
find lim(x->0)f(x)=?

I have done all my hits all failed!!!
can you please tell me how to solve it???

2. Apr 12, 2012

### HallsofIvy

Well, tell us what you have tried so we will know what hints will help.

3. Apr 12, 2012

### vkash

i have reached to this answer!!! is it correct..

let y = [x^(sin(2x)] cos(2x)
ln(y)= sin(2x)*ln(x)+ln(cos(2x))
lim(x->0) sin(2x)*ln(x)+ln(cos(2x))
lim(x->0) sin(2x)*ln(x) //(as ln(cos(0))=ln(1)=0
lim(x->0)2sin(x)ln(x)
lim(x->0)2 ln(x)/cosec(x)
L Hopital
lim(x->0)-2/(xcosec(x)cot(x))
lim(x->0)-2sin2(x)/xcos(x)
lim(x->0)-2sin2(x)/x
once again L hopital
lim(x->0)-4sin(x)cos(x)+2sin3(x)
=0.................(area all steps correct)
ln(y)=0
y=1
;lim(x->0)f(x)=2y=2
is it correct. I doesn't have marking scheme or any thing like this so i can't tell what's correct answer..

Last edited: Apr 12, 2012
4. Apr 12, 2012

### SammyS

Staff Emeritus
The correct answer is 2.

5. Apr 12, 2012

### vkash

you mean my answer is correct???????
OR
you have already done this question and saying me that answer is 2??

6. Apr 12, 2012

### SammyS

Staff Emeritus
The answer is correct.

The step that goes from
lim(x->0) sin(2x)*ln(x)​
to
lim(x->0)2sin(x)ln(x)​
is not correct, if you mean that sin(2x)ln(x) = 2sin(x)ln(x) .

7. Apr 12, 2012

### Dick

I don't think that's right.

8. Apr 16, 2012

### vkash

are nahi yar!!
no i have just skipped a step..
lim(x->0)sin(2x)ln(x) =lim(x->0)2sin(x)cos(x)ln(x)
cos(0)=1
so
lim(x->0)sin(2x)ln(x) =lim(x->0)2sin(x)cos(x)ln(x)=lim(x->0)2sin(x)ln(x)

so what do you think about it..

9. Apr 16, 2012

### Dick

Ok so far. I'd double check if you still think the limit is 2.

10. Apr 16, 2012

### SammyS

Staff Emeritus
I'm pretty sure that $\displaystyle \lim_{x\to0^+}2\,x^{\sin(2x)} \cos(2x)=2\,.$

11. Apr 16, 2012

### Dick

Yeah, you're right. I've been missing the initial '2' somehow. Sorry.

12. Apr 16, 2012

### vkash

thanks to all of you for helping me...