IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

cuteprincess

find lim x $\rightarrow$ $\pi$ /2 log sinx / ( $\pi$ -2x)²

allamraju

Ans is -1/8.

ashish_banga

-1/8 use L hospital

allamraju

Use l'hospital rule twice as it is in the form of 0/0.

So,lim cotx/2(pi-2x)(-2)=(-1/4)lim-cosec²x/-2=-1/8 as cosecx tends to 1 as x tends to pi/2

cuteprincess

Ya..
The answers are rite
But is there ne short-cut method to cross check the answer???...

cuteprincess

A re pls nebody answer na...

allamraju

What do u mean by short-cuts?The ans is itself short,only two steps.

cuteprincess

Hey I just meant
Is there any method of cross checking..??

allamraju

No,I don't think they are any.For such problems,Apply L'hospital rule repeatedly till it reduces to the form whose limit exists finitely.

budokai_tenkaichi_returns

no u cant check ur solutions in limits ..

go4it

use lhosp. and solve....ans is -1/8

nunoxic

Cross Check by solving "ijjat se" :D
put x = pi/2 + h

u get denominator as
4h^2
numerator as log cos x

lim h tends to 0 log cos x / 4h^2

= log cos^2 x / 8h^2
put cos^2 x = 1 - sin^2 x

log (1+x) = x (x tends to 0)

thus :
- sin^2 x / 8 h^2
-1/8

but as the brains above suggested
L hopital rox but since u wanted another method i provided u

cuteprincess

how is the numerator log(cosx)..??

nd x tends to pi/2...

pls explain..

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Ans is -1/8.

Use l'hospital rule twice as it is in the form of 0/0.

So,lim cotx/2(pi-2x)(-2)=(-1/4)lim-cosec2x/-2=-1/8 as cosecx tends to 1 as x tends to pi/2

What do u mean by short-cuts?The ans is itself short,only two steps.

No,I don't think they are any.For such problems,Apply L'hospital rule repeatedly till it reduces to the form whose limit exists finitely.

So,lim cotx/2(pi-2x)(-2)=(-1/4)lim-cosec²x/-2=-1/8 as cosecx tends to 1 as x tends to pi/2