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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 May 2008 20:43:54 IST
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limit...............
limit x tending to  /4 2 - (Cosx+Sinx)*3 /1-Sin2x
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16 May 2008 20:55:13 IST
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where the ques?????????????
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"Imagination is more important than knowledge."
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17 May 2008 16:32:12 IST
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hey man i think its easy
lim 2root2 - 3*(cos x +sin x)/1-sin2x
just multiply and divide the 2nd term by 2x we get
lim 2root2 - 3*{(cos x +sinx)/2x}/(1/2x)-(sin2x/2x)
sin x/x =1 as x tends t0 0
ttherefore put sin2x/2x=1 and solving we get
ans= - 2root2( +1)
hope u got it...
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Ans is 3/rt2.Since it is pf the form 0/0,Apply L'hospital's rule,then
-3(cosx+sinx)2(cosx-sinx)/-2cos2x=(3/2)[(cosx+sinx)2(cosx-sinx)/cos2x-sin2x].By cancelling (cosx-sinx) and one (cosx+sinx),it becomes (3/2)(cosx+sinx) which tends to 3/rt2 as x tends to pi/4
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MAKING A MISTAKE IS HUMAN BUT REPEATING IT IS IDIOTIC. |
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17 May 2008 16:42:41 IST
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Limit is x tending to pi/4..
I think question should be (cosx+sinx)*2.. Then the answer comes out to be 2root2.. Nahi to nahi ho raha..
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17 May 2008 16:43:32 IST
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It's not of the form zero by zero
Numerator's not zero in current form..
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17 May 2008 16:44:30 IST
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hey allamraju it is not of 0/0 form i think u r mistaken
and i think that my soln is correct.
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17 May 2008 16:47:05 IST
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What I think is it is (cosx+sinx)3 and not 3(cosx+sinx) and hence posted the soln.
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MAKING A MISTAKE IS HUMAN BUT REPEATING IT IS IDIOTIC. |
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17 May 2008 16:50:29 IST
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Oh k.. That makes a lot more sense.. Then you're right. 
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17 May 2008 16:50:44 IST
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Oh k.. That makes a lot more sense.. Then you're right.
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17 May 2008 16:54:43 IST
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ans is 5/sqrt2 if 2sqrt2 is independent of the othr frac
othr wise allamraju is correct
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17 May 2008 16:57:36 IST
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how can u plz tell the method?
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17 May 2008 16:58:12 IST
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if u asked me
ive edited it in my prev post
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17 May 2008 17:07:02 IST
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hey man i m not getting it can u tell the soln.?
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17 May 2008 17:17:43 IST
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if 2sqrt is included in fraction then allam rajus ans is right
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