# 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: 39,682 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..

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P: 7,819
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..
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: 7,819
 Quote by vkash you mean my answer is correct??????? OR you have already done this question and saying me that answer is 2??

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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P: 25,235
 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..
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P: 25,235
 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: 7,819 I'm pretty sure that $\displaystyle \lim_{x\to0^+}2\,x^{\sin(2x)} \cos(2x)=2\,.$
 Quote by SammyS I'm pretty sure that $\displaystyle \lim_{x\to0^+}2\,x^{\sin(2x)} \cos(2x)=2\,.$