# find out the limit of the following function...

by vkash
Tags: function, limit
 P: 318 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???
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,900 Well, tell us what you have tried so we will know what hints will help.
P: 318
 Quote by HallsofIvy 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..

Emeritus
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P: 7,418

## find out the limit of the following function...

 Quote by 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..
The correct answer is 2.
P: 318
 Quote by SammyS 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??
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P: 7,418
 Quote by vkash 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) .
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Thanks
P: 25,175
 Quote by SammyS The correct answer is 2.
I don't think that's right.
P: 318
 Quote by SammyS The answer is correct. The step that goes fromlim(x->0) sin(2x)*ln(x)tolim(x->0)2sin(x)ln(x)is not correct, if you mean that sin(2x)ln(x) = 2sin(x)ln(x) .
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..
HW Helper
Thanks
P: 25,175
 Quote by 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..
Ok so far. I'd double check if you still think the limit is 2.
 Emeritus Sci Advisor HW Helper PF Gold P: 7,418 I'm pretty sure that $\displaystyle \lim_{x\to0^+}2\,x^{\sin(2x)} \cos(2x)=2\,.$
 Quote by SammyS I'm pretty sure that $\displaystyle \lim_{x\to0^+}2\,x^{\sin(2x)} \cos(2x)=2\,.$