Question

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

Answer 1

`lim_(x->5^-) (x-5)/(x^2-25) = 1/10`

Explanation:

`lim_(x->5^-) (x-5)/(x^2-25) = lim_(x->5^-) color(red)(cancel(color(black)((x-5))))/(color(red)(cancel(color(black)((x-5))))(x+5))`

`color(white)(lim_(x->5^-) (x-5)/(x^2-25)) = lim_(x->5^-) 1/(x+5)`

`color(white)(lim_(x->5^-) (x-5)/(x^2-25)) = 1/(5+5)`

`color(white)(lim_(x->5^-) (x-5)/(x^2-25)) = 1/10`

Answer 2

`1/10`

Explanation:

Substituting `5` in the given expression results to an indeterminate solution:

`lim_(x->5^-)(x-5)/(x^2-25)`

`=(5-5)/(5^2-25)`

`=0/0" "` Indeterminate solution.

let us think about another way to find the limit.

Here, Factorizing `x^2 - 25` is the best way.

Factorization `x^2 -25" "`is computed by applying the

polynomial identity.

`color(blue)((a^2-b^2)=(a-b)(a+b))`

`color(blue)(x^2 - 25=x^2 - 5^2=(x-5)(x+5))`

`lim_(x->5^-)(x-5)/(x^2-25)`

`=lim_(x->5^-)cancel(x-5)/color(blue)(cancel(x-5)(x+5))`

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

`=1/(5 + 5)`

`=1/10`