lim x→8 {[1+(1+x)^1/2 ]^1/2 -2}/x-8

Pls answer. sorry for this uncomfortable font

Hi Saideep.

Multiply Nr and Dr by the conjugate of the Nr Twice.

When you do it for the first time, the Nr will be (1+x)1/2-3.

And the next time when you do it, the Nr will become x-8.

And this will get cancelled with the x-8, in the Dr, and then you can put x=8, in the resulting expression.

The resulting expression would be 1/{[1+(1+x)^1/2 ]^1/2 + 2}*{(1+x)^1/2 +3}.

Put x=8 in this expression and you will get the limit as, 1/24.

Hope that helps.

All the best.

Regards,

Ashwin (IIT Madras).

You can directly differnsiate nr and dr sepratly twice and get answer

x->8

reduces to lim  d/dx [1/{{1+(1+x)^1/2}^1/2} / x-8]

               x->8

=lim 1 / [2{{1+(1+x)^1/2}^1/2} * 1  / [2{(1+x)^1/2}

x->8

Upon putting the value of x=8,we get the required limit=1/24.