IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

thedumbheadwithnobrain

Answer is 0,posting solution in next post plz wait

thedumbheadwithnobrain

$\mathop{\lim}\limits_{{x} \to {0}}\frac{x\;cosx}{x-sinx}$

differentiating wrt x two times by LH rule

$\mathop{\lim}\limits_{{x} \to {0}}\frac{-2sinx-xcosx}{sinx}=-2-\frac{cosx}{\frac{sinx}{x}}=-2-\frac{1}{1}=-3$

sandeepramesh

wrong! Your denominator is wrong after twice differentiating!

allamraju

I think the ans is -3,After differentiating twice by LH rule,we get,

Lim(-2sinx-xcosx/sinx)=Lim (-2)-[cosx/(sinx/x)] which tends to -2-(1/1)=-3 as x tends to 0.

somyekathait

sorry !! that was my fault

xcosx / x-sinx

= 1/(secx- tanx/x )

= 1/infinite =0

thedumbheadwithnobrain

@somye

u wil get $\mathop{\lim}\limits_{{x} \to {0}}\frac{-xsinx+cosx}{1-cosx}$ after diffrentiating once

allamraju

Are yaar..What is going on!.Somyekathait,the denominator must be 1-cosx and not -cosx.Check it.

akku

@

somyekathait diff of x is 1 nt 0

@

allamraju ;

thedumbheadwithnobrain;

Can sme1 explain me why????? every1 is diff the fn fr the 2nd time when its 1/0 and NOT 0/0 form .How can u apply LH rule ????

somyekathait

sorry !! that was my fault

xcosx / x-sinx

= 1/(secx- tanx/x )

= 1/infinite =0

thedumbheadwithnobrain

u r again wrong

$\mathop{\lim}\limits_{{x} \to {0}}\frac{1}{secx-\frac{tanx}{x}}=\frac{1}{1-1}=\frac{1}{0}$ which is not defined

akku

How does the infinite cme in den
as x approaches 0 secx--->1 tanx/x--->1 so den--->0

allamraju

Thanks Akku,I didn't observe that.When sandeep posted that there was an error in 2nd derivative in thumbhead's ans,I corrected that and evaluated the limit.He,later edited his message..Sorry for the mistake.

sandeepramesh

yes, my first wrong was bcos the answer was wrong and the 2nd wrong was because the wrong answer's wrong method was wrong!

iitjee08aspirant

is the answer not defined since it is of form 1/0.....

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I think the ans is -3,After differentiating twice by LH rule,we get,

Lim(-2sinx-xcosx/sinx)=Lim (-2)-[cosx/(sinx/x)] which tends to -2-(1/1)=-3 as x tends to 0.

Are yaar..What is going on!.Somyekathait,the denominator must be 1-cosx and not -cosx.Check it.

Thanks Akku,I didn't observe that.When sandeep posted that there was an error in 2nd derivative in thumbhead's ans,I corrected that and evaluated the limit.He,later edited his message..Sorry for the mistake.