Question

Question #89992

Answer 1

The limit is `0`

Explanation:

For `h > 0`, we have `1+h > 1`.

So, for `h > 0`, we have `f(1+h) = ((1+h)-1)^2`.

This simplifies to `f(1+h) = h^2` `" "` (for `h > 0`).

Therefore,

`lim_(hrarr0^+)f(1+h) - lim_(hrarr0^+)h^2 = 0`