IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

mastermind890

_{[ x]}

_{[ 0]} (sin x)^x

dilip

I think ans. is 1

mastermind890

actually a frnd of mine asked me dis question...he said the answer was 0....but plz show me how did you get 1....

ramyadiamond

Let L=_{[ x]}_{[ 0]} (sinx)^x

take log on both sides,

log L = _{[ x][0 ]} x.log(sinx)

(i'm not typing limit all the time, ok? its just there in every step)

= log(sinx)/(1/x)

now applying L-Hospital's rule,

=(cosx/sinx)/(-1/x²)

= - (x.cosx)/(sinx/x)

the limit in the denominator becomes 1 on applying limit, and the numerator becomes 0

hence, logL=0

L=e⁰

Hence, L=1.

So, dilip is right.

karthik2007

For such limit problems where the power is raised to a variable, taking log should solve it most of the times.

gaurav_5198

No ur frnd told u wrong. It's ans is 1. Here is the soln.

A = Lim [x->0] (sinx)^x

Take log on both sides

log A = Lim [x->0] x log sinx

= Lim [x->0] ( log sinx ) / ( 1/x)

Apply L'Hospital rule ,

= Lim [x->0] ( cosx/sinx ) / ( -1/x²)

= Lim [x->0] ( cosx . x . x / sinx )

= Lim [x->0] ( cosx . x ) [ Lim [x->0] (sinx)/x = 1 ]

= 1 * 0 = 0

Now , A = e⁰ = 1 ( -- Ans. )

mastermind890

limit doesnt exist....

johri_anshuman

RHL=1
LHL does not exist

so the reqd limit does not exist