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Find out the limit of the following function...

by vkash
Tags: function, limit
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vkash
#1
Apr12-12, 08:17 AM
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???
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HallsofIvy
#2
Apr12-12, 08:52 AM
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Well, tell us what you have tried so we will know what hints will help.
vkash
#3
Apr12-12, 10:12 AM
P: 318
Quote Quote by HallsofIvy View Post
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
#4
Apr12-12, 11:29 AM
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Find out the limit of the following function...

Quote Quote by vkash View Post
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
#5
Apr12-12, 08:25 PM
P: 318
Quote Quote by SammyS View Post
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
#6
Apr12-12, 10:44 PM
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Quote Quote by vkash View Post
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
#7
Apr12-12, 10:57 PM
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Quote Quote by SammyS View Post
The correct answer is 2.
I don't think that's right.
vkash
#8
Apr16-12, 09:48 AM
P: 318
Quote Quote by SammyS View Post
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) .
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
#9
Apr16-12, 09:57 AM
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Quote Quote by vkash View Post
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
#10
Apr16-12, 12:42 PM
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I'm pretty sure that [itex]\displaystyle \lim_{x\to0^+}2\,x^{\sin(2x)} \cos(2x)=2\,.[/itex]
Dick
#11
Apr16-12, 12:49 PM
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Quote Quote by SammyS View Post
I'm pretty sure that [itex]\displaystyle \lim_{x\to0^+}2\,x^{\sin(2x)} \cos(2x)=2\,.[/itex]
Yeah, you're right. I've been missing the initial '2' somehow. Sorry.
vkash
#12
Apr16-12, 02:03 PM
P: 318
thanks to all of you for helping me...


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