# How do you find the limit of lnt/sqrt(1-t) as t->1^-?

Jan 12, 2017

${\lim}_{x \to {1}^{-}} \ln \frac{t}{\sqrt{1 - t}} = 0$

#### Explanation:

We have that:

${\lim}_{x \to {1}^{-}} \ln t = 0$

${\lim}_{x \to {1}^{-}} \sqrt{1 - t} = 0$

so that the limit:

${\lim}_{x \to {1}^{-}} \ln \frac{t}{\sqrt{1 - t}}$

is in the indeterminate form $\frac{0}{0}$

We can then use l'Hospital's rule:

${\lim}_{x \to {1}^{-}} \ln \frac{t}{\sqrt{1 - t}} = {\lim}_{x \to {1}^{-}} \frac{\frac{d}{\mathrm{dt}} \ln t}{\frac{d}{\mathrm{dt}} \sqrt{1 - t}} = {\lim}_{x \to {1}^{-}} \frac{\frac{1}{t}}{- \frac{1}{2} \frac{1}{\sqrt{1 - t}}} = {\lim}_{x \to {1}^{-}} - 2 \frac{\sqrt{1 - t}}{t} = 0$