Question

How do you find the limit of `(x-5)/(x^2-25)` as `x->5`?

Answer 1

`1/10`

Explanation:

`(x-5)/(x^2-5)=(x-5)/((x-5)(x+5))=1/(x+5), " for each " x!=5`

so

`lim_(x->5) f(x) = lim_(x->5) 1/(x+5) = 1/10`