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

Question

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

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Answer 1 · 2017-03-24T23:48:36

Please see below.

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

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.