Let # f : RR rarr RR#, be a function and let #x in RR#. We say that the

function #f# is **Differentiable** at #x# , if, the following **Limit**

exists # : lim_(trarrx) (f(t)-f(x))/(t-x), where, t in RR, t!=x.#

Also, if the above Limit exists, it is called the **Derivative** of #f# at

#x#, and, is denoted by #f'(x).#

We have, #f(x)=(2x+1)/(x-2), x in RR-{2}#

#rArr f(t)=(2t+1)/(t-2), t in RR-{2}, tnex#

#:. f(t)-f(x)=(2t+1)/(t-2)-(2x+1)/(x-2)#

#={(2t+1)(x-2)-(2x+1)(t-2)}/((t-2)(x-2))#

#={(2tx-4t+x-2)-(2tx-4x+t-2)}/((t-2)(x-2))#

#=(-5t+5x)/((t-2)(x-2))=(-5(t-x))/((t-2)(x-2))#

Since, #t!=x,# we have, #(f(t)-f(x))/(t-x)=-5/((t-2)(x-2))#

Now, #lim_(trarrx) (f(t)-f(x))/(t-x)=lim_(trarrx) -5/((t-2)(x-2))#

#=-5/((x-2)(x-2))=-5/(x-2)^2#.

Thus, we find that, the **Limit** in question exists, and so, the given

fun. #f# is differentiable at #x in RR-{2}, and, f'(x)=-5/(x-2)^2, x in RR-{2}.#

Enjoy maths.!