Question

Find the limit as x approaches 3 from the right `ln(x^2-9)`?

I keep on getting oo, however the answer is -oo. Isnt `(x^2-9)` going to give a postive

Answer 1

Please see below.

Explanation:

For `u > 1`, we get `lnu >0`, but for `0 < u < 1`, we have `ln u < 0`.

So `lim_(urarr0^+) lnu = -oo`

As `xrarr3^+`, we have `(x^2 - 9) rarr 0^+`

Therefore, `lim_(xrarr3^+)ln(x^2-9) = -oo`

A little more to think about

For `x = sqrt10 ~~ 3.1623`, we have `x^2 -9 = 1`, so `ln(x^2 - 9) = ln1 = 0`

As `x` continues toward `3` from the right, we have `x^2` is between `9` and `10`, so `(x^2 - 9)` is between `0` and `1` and its ln is negative.