using L'Hôpital's Rule on an arctan This is part of a larger problem, but now I'm trying to find 
$$\lim_{n\to\infty}\left(\frac12 + \frac1{\pi}\arctan\left(\frac{nx}{t}\right)\right)^n.$$
I think I have to use L'Hôpital's, but I'm not sure how.
 A: An indeterminate of the form $1^{\infty}$ can be solved through L'Hopital's Rule and the use of logarithms.
First, note that $a^b = \mathrm{exp}(b\ln a)$. Since the exponential function is continuous, the limit of an exponential is the exponential of the limit, so we can compute the limit of $a^b$ by computing the limit of $b\ln a$ instead (and then taking the exponential of that).
If we have that $a^b$ is an indeterminate of the form $1^{\infty}$ (that is, $a\to 1$ and $b\to\infty$), then $b\ln(a)$ is an indeterminate of type $\infty\times 0$ (since $b\to\infty$ and $\ln(a)\to-\infty$). You can usually apply L'Hopital's Rule to indeterminates of the form $0\times\infty$ by moving one of the factors to the denominator, so it becomes either a $\frac{0}{0}$ indeterminate, or a $\frac{\infty}{\infty}$ indeterminate.
To put all of this to work in this example, we note that as $n\to\infty$, $\arctan(nx/t)\to \frac{\pi}{2}$, so this is indeed an indeterminate of type $1^{\infty}$. We will first try to find
$$\lim_{n\to\infty} n\ln\left(\frac{1}{2}+\frac{1}{\pi}\arctan\left(\frac{nx}{t}\right)\right)$$
and then take the exponential of this limit. To compute this limit, we rewrite it and use L'Hopital's Rule for a $\frac{0}{0}$-indeterminate:
$$
\lim_{n\to\infty}n\ln\left(\frac{1}{2}+\frac{1}{\pi}\arctan\frac{nx}{t}\right) = \lim_{n\to\infty}\frac{\ln\left(\frac{1}{2} + \frac{1}{\pi}\arctan\frac{nx}{t}\right)}{\frac{1}{n}}.$$
The derivative of the numerator is
$$\left(\frac{1}{\frac{1}{2}+\frac{1}{\pi}\arctan\frac{nx}{t}}\right)\left(\frac{x}{t\pi}\right)\left(\frac{1}{1 + \frac{n^2x^2}{t^2}}\right)
= \left(\frac{1}{\frac{1}{2}+\frac{1}{\pi}\arctan\frac{nx}{t}}\right)\left(\frac{x}{t\pi}\right)\left(\frac{t^2}{t^2 + n^2x^2}\right)
$$
The derivative of the denominator is simply $-\frac{1}{n^2}$. So the limit we want to compute is:
$$\begin{align*}
\lim_{n\to\infty}\frac{\ln\left(\frac{1}{2} + \frac{1}{\pi}\arctan\frac{nx}{t}\right)}{\frac{1}{n}} &\stackrel{\mathrm{L'H}}{=} \lim_{n\to\infty}\left(\frac{1}{\frac{1}{2} + \frac{1}{\pi}\arctan\frac{nx}{t}}\right)\left(\frac{x}{t\pi}\right)\left(\frac{-t^2n^2}{t^2+n^2x^2}\right)\\
&=\frac{1}{1}\times\frac{x}{t\pi}\times\frac{-t^2}{x^2}\\
&= -\frac{t}{x\pi}.
\end{align*}$$
And so our original limit is the exponential of this,
$$\lim_{n\to\infty}\left(\frac{1}{2}+\frac{1}{\pi}\arctan\frac{nx}{t}\right)^n = \mathrm{exp}\left(-\frac{t}{x\pi}\right).$$
A: Since someone else already answered in full, I'll turn my hint into a full answer.
Notice that $\arctan(x) = {\pi \over 2} - \arctan({1 \over x})$. Therefore we have
$$({1 \over 2} + {1 \over \pi}\arctan({nx \over t}))^n = (1 - {1 \over \pi}\arctan({t \over nx}))^n$$
It's easier to take logs of this expression, find the limit as $n$ goes to $\infty$ and then take $e$ to the answer. So our goal is to find
$$\lim_{n \rightarrow \infty} n\ln(1 - {1 \over \pi}\arctan({t \over nx}))$$
Note that $\ln(1 - y) = -y + o(y^2)$ and $\arctan(z) = z + o(z^2)$. So we have
$$\ln(1 - {1 \over \pi}\arctan({t \over nx})) = -{1 \over \pi}\arctan({t \over nx}) + 
O(1/n^2)$$
$$ = -{1 \over \pi} {t \over nx} + O({1 \over n^2})$$
Multiplying this by $n$ and taking limits as $n$ goes to infinity, we obtain ${\displaystyle -{1 \over \pi}{t \over x}}$. Taking $e$ to this, we see that the answer is ${\displaystyle e^{-{t \over \pi x}}}$
