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**Paperwings** Evaluate

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

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

$\displaystyle \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

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

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 $\displaystyle \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.