Question

Question #f063f

Answer 1

`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`