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![[Post New]](/templates/default/images/icon_minipost_new.gif) 25 Feb 2008 20:24:45 IST
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Solve:
LIMx-->0 (x cosx+sin x)/(x2 +tan 2x)
The answer is 11/6.
Please show the steps as well...
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Be not afraid of growing slowly.
Be afraid only of standing still.
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25 Feb 2008 20:41:17 IST
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Please PPL RATES R READY TO BE GIVEN AWAY...
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Be not afraid of growing slowly.
Be afraid only of standing still.
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25 Feb 2008 20:58:00 IST
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hey!!!!!!!!!!!!!!!
but on applying l hospitals rule twice ans coming is 0?????
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FAILURE IS NOT FALLING IN LIFE BUT NOT RISING AGAIN AFTER FALLING!!!!!!
I LIKE WAVES NOT BECAUSE THEY RISE AND FALL..
BUT BECAUSE EVERYTIME THEY FALL THEY RISE AGAIN!!!!!!!
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25 Feb 2008 21:10:24 IST
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hey i solvd it widout applying Lhospital........bt evn i m gtng d ans as 0....... plz sum1 clarify d doubt yar
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25 Feb 2008 22:36:19 IST
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truly it comes =0
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25 Feb 2008 23:01:26 IST
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Since numerator is 0 and denominator is 0 we can apply lopital rule Applying lopital rule once we get (cosx-xsinx + cosx)/(2x+sec2x) now it is no more in indeterminate form so applying the limit (1-0+1)/(0+1) =2 i dont see how the answer can be 0 or for that matter 11/6
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25 Feb 2008 23:13:18 IST
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Sboosy , u r wrong man!!!!!!! Here lim (xcosx + sinx)/x2+tan2x x-0 Differentiating u get lim (2cosx-xsinx)/2x+2tanxsec2x x-0 Put x=0 2-0 / 0+0 = 
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25 Feb 2008 23:15:17 IST
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sorry raulrag
i copied the question wrong
i took it to b tanx in the denominator
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25 Feb 2008 23:37:53 IST
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no problem!!
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26 Feb 2008 08:21:34 IST
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Please experts,solve this problem...
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Be not afraid of growing slowly.
Be afraid only of standing still.
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26 Feb 2008 10:49:26 IST
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![=\lim_{x->0}\frac{1}{2}[ \frac{2cosx-xsinx}{x+\frac{sinx}{(cosx)^3}} ]](http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/f/0/6/f0652e70d8e4847561315f8768d3b586eb762046.gif)
![=\lim_{x->0}\frac{1}{2}(cosx)^3[ \frac{2cosx-xsinx}{x(cosx)^3+sinx} ]](http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/f/7/a/f7a5bd4998b13a8b3f998d324d318379f47adff1.gif)
![=\lim_{x->0}\frac{1}{2}[ \frac{2}{x(cosx)^3+sinx} ]](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/5/6/5/5659b345225d9f17073da802e4de8d928b34963e.gif)
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26 Feb 2008 11:38:26 IST
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No,dude that is not the answer......
It is a question from 11th std R.D.Sharma..........
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Be not afraid of growing slowly.
Be afraid only of standing still.
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26 Feb 2008 12:00:02 IST
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Re:Trigonometric limits...
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26 Feb 2008 15:10:15 IST
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Thank u ppl for the hard effort from u'r side...I think the answer given in the back is wrong.I think it is infinity............Thanks snipy
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Be afraid only of standing still.
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