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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 May 2008 14:45:49 IST
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limits problem
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3 May 2008 14:47:36 IST
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lt tanx-sinx/x
here x tends to 0
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3 May 2008 15:05:35 IST
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answer =-1
as lim x->0 tanx=0
and lim x->0 sinx/x=1
so 0-1=-1
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3 May 2008 15:16:02 IST
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divided by x fr da whole thing
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3 May 2008 15:16:33 IST
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0
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Nitwit Blubber Odment Tweak
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3 May 2008 15:17:20 IST
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then the ans. is 0
simple .....apply l'hosp. rule .....u will get 0 :)
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3 May 2008 15:17:47 IST
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sry yar its divide by (x)cube sry
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3 May 2008 15:22:31 IST
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its 1/6 ???
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* "A Wish will not change anything, however a Decision will change everything"
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3 May 2008 15:23:40 IST
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nope da ans is 1/2
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3 May 2008 15:26:31 IST
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no the ans. is 1/2
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3 May 2008 15:27:18 IST
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how did u do it?????
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3 May 2008 15:30:23 IST
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yaa got it...
apply L'hosp. rule
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(\frac{1-cosx}{{x}^{2}})\\=%20(1)\mathop{\lim}\limits_{{x}%20\to%20{0}}%20(\frac{1}{2})\frac{2[{sin(\frac{x}{2})}]^{2}}{[{\frac{x}{2}}]^{2}})
= 1 / 2
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3 May 2008 15:33:28 IST
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thanx
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3 May 2008 15:34:04 IST
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tanx-sinx=sinx(1-cosx)/cosx = sinx(2sin^2(x/2))/cosx
as x tends to zero, you can take sinx= x, sin x/2=x/2, cos x=1
so, (x.2.x/2.x/2.1)/x^3 =1/2
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