Question #f063f

1 Answer
Aug 31, 2017

#lim_(x->p) (1/x-1/p)/(x-p) = -1/p^2#

Explanation:

As #x->p#, #1/x# will also approach #1/p#

Both #x-p# and #1/x-1/p# would thus approach 0

Hence, you will end up with the indeterminate form #0/0#

Here, you would be able to use L'Hôpital's rule

Find #d/dx(1/x-1/p)# and #d/dx(x-p)#

Since p is a constant, #d/dx(1/x-1/p)=-1/x^2# and #d/dx(x-p)=1#

Hence, by L'Hôpital's rule,

#lim_(x->p) (1/x-1/p)/(x-p) = lim_(x->p) (-1/x^2)/(1) = -1/p^2#