How to find $ \lim_{T \to \infty} \frac{1}{1-e^{-\theta/T}} $? I believe this should be $\frac{T}{\theta}$, but I am not sure how to get there. 
So far I have tried rational fractions techniques, e.g. breaking it into $1 + \dfrac{1}{e^{\theta/T}-1}$ and also trying to make L'Hopital's Rule applicable by changing the fraction to be $\dfrac{1-e^{-\theta/T}}{\left(1-e^{-\theta/T}\right)^2}$, however neither of these have yielded progress.
 A: When $T \to + \infty$, $\frac{-\theta}{T}$ tends to zero regardless of $\theta$. So the limit is $\infty$ as the numerator is $1$ and the denominator tends to zero. 
If $\theta > 0$
$$
\lim_{T \to +\infty} \frac{1}{1-e^{-\theta/T}} = \frac{1}{0^+} = +\infty
$$
and, if $\theta < 0$,
$$
\lim_{T \to +\infty} \frac{1}{1-e^{-\theta/T}} = \frac{1}{0^-} = -\infty
$$
[Edit]
If you consider $T/\theta$ instead of $\theta/T$ you still have two possibilities:
If $\theta >0$,
$$
\lim_{T\to +\infty}\frac{1}{1-e^{-T/\theta}}= \frac{1}{1-0} = 1
$$
If $\theta < 0$,
$$
\lim_{T\to +\infty}\frac{1}{1-e^{-T/\theta}}= \frac{1}{-\infty} = 0.
$$
A: NOTE L'Hopital's Rule does not apply to limits of the form $1/0$.
HINT For small $x$, you have $e^x \approx 1 +x,$ so
$$
\frac{1}{1-e^{-x}} \approx \frac{1}{1 - (1+x)} = \frac{-1}{x}.
$$
Can you finish?
A: Let


*

*$u=\theta/T\to 0$
then
$$\lim_{T \to +\infty} \frac{1}{1-e^{-\theta/T}} = \frac{1}{0^+} =\lim_{u \to 0} \frac{1}{1-e^{u}}=\lim_{u \to 0} \left(\frac1u \cdot\frac{u}{1-e^{u}}\right)$$
and recall that by standard limit
$$\frac{u}{1-e^{u}}=-\frac{1}{\frac{e^{u}-1}u}\to -1$$
