Question

If the function f is continuous at = 2, then find f(2) where f(x) = `(x^5 - 32)/(x - 2)`, for ≠ 2.

Answer 1

Consider, 

`lim_(x ->2) f(x) = lim_(x->2) [(x^5 - 32)/(x - 2)]`

`lim_(x->2) [(x^5 - 2^5)/(x - 2)]`

= 5 (2)^5-1

`(lim_(x->a) (x^n - a^n)/(x -a) = na^n-1)`

= 80

Since f is continuous at x = 2

`lim_(x->2)` f(x) = f(2)

f(2) = 80