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Find out the limit of the following function... 
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#1
Apr1212, 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??? 


#2
Apr1212, 08:52 AM

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



#3
Apr1212, 10:12 AM

P: 318

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)2sin^{2}(x)/xcos(x) lim(x>0)2sin^{2}(x)/x once again L hopital lim(x>0)4sin(x)cos(x)+2sin^{3}(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.. 


#4
Apr1212, 11:29 AM

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Find out the limit of the following function...



#5
Apr1212, 08:25 PM

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


#6
Apr1212, 10:44 PM

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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
Apr1212, 10:57 PM

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#8
Apr1612, 09:48 AM

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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
Apr1612, 09:57 AM

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#10
Apr1612, 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]



#11
Apr1612, 12:49 PM

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#12
Apr1612, 02:03 PM

P: 318

thanks to all of you for helping me...



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