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