# Thread: [SOLVED] Natural Log Limit

1. ## [SOLVED] Natural Log Limit

Evaluate
$\lim_{t\to-5} \ln(t+5)^2$

Since $\ln(0)^2$ is undefined, I used L'Hopital's Rule twice to get

$\lim_{t\to-5} \ln(t+5)^2 = \lim_{t\to-5}\frac{2 \ln(t+5)}{t+5} = \frac{0}{0}$

This gives an indeterminate case as well. So I used L'Hopital's Rule once again to get

$\lim_{t\to-5}\frac{2 \ln(x+5)-1}{(x+5)^2} = \frac{0}{0}$

which brings be back to an indeterminate case because of the x+5 in the denominator irregardless of how many times I take the derivative.

I know that $\lim_{x\to0+} \ln(x) \to -\infty$, but don't know how to apply it here.

Graphing it on my calculator, I see that the limit goes to positive infinity.

2. Originally Posted by Paperwings
Evaluate
$\lim_{t\to-5} \ln(t+5)^2$

Since $\ln(0)^2$ is undefined, I used L'Hopital's Rule twice to get

$\lim_{t\to-5} \ln(t+5)^2 = \lim_{t\to-5}\frac{2 \ln(t+5)}{t+5} = \frac{0}{0}$

This gives an indeterminate case as well. So I used L'Hopital's Rule once again to get

$\lim_{t\to-5}\frac{2 \ln(x+5)-1}{(x+5)^2} = \frac{0}{0}$

which brings be back to an indeterminate case because of the x+5 in the denominator irregardless of how many times I take the derivative.

I know that $\lim_{x\to0+} \ln(x) \to -\infty$, but don't know how to apply it here.

Graphing it on my calculator, I see that the limit goes to positive infinity.
I am not 100% sure about this, but I think that ln(0)^2 does not count as an indeterminate form. There are a specific set of situations and they all involve more than one term. See this Wiki page on it:

Indeterminate form - Wikipedia, the free encyclopedia

Now as for your limit, remember that ln(x) is not defined for $x \le 0$. You have a horizontal shift from ln(x) but the concept still applies. When you graph your equation you should not see anything from x=-5 and below, thus the limit from the left hand side DNE. That is not a rigorous proof of course but it sounds like you more so need an explanation than proof.

3. Thank you. The limit doesn't exist because the one-sided limit from the left doesn't exist.