Find out the limit of the following function

[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?

 

[HallsofIvy]

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

 

[vkash]

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

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..

 

[SammyS]

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..

The correct answer is 2.

 

[vkash]

The correct answer is 2.

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

 

[SammyS]

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

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) .

 

[Dick]

The correct answer is 2.

I don't think that's right.

 

[vkash]

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)[/color] .

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..

 

[Dick]

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..

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

 

[SammyS]

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

 

[Dick]

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

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

 

[vkash]

thanks to all of you for helping me...