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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.

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    Yes, you can just type standard LaTeX, enclosed in `$` for in-line formulas and in `$$` for displayed formulas.2011-04-27
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    You don't say the limit as what goes where, but I assume it's the limit as $n\to\infty$.2011-04-27
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    Limit as what is going where?2011-04-27
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    yes, limit as n goes to infinity2011-04-27
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    Is it perhaps $$ (1/2 - (1/\pi) arctan(nx/t))^n $$ instead?2011-04-27
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    Hint: Let $u_n=\left(\frac{1}{2} + \frac{1}{\pi}\arctan\left(\frac{nx}{t}\right)\right)-1$. Since $\lim_{n \to \infty} u_n=0$, then $\lim_{n \to \infty} (1+u_n)^\frac{1}{u_n}= e$. Then use L'H..... You can also threat it as a general $1^\infty$ L'H limit.2011-04-27
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    Do you mean use L'H on $u_n$?2011-04-27
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    I got something like $\exp(\pm t/x\pi)$ by rewriting the arctangent as the arc tangent of something that goes to zero, and then using the series for arctangent around zero.2011-04-27
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    @mixedmath: Instead of what?2011-04-27
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    @Carl: I got the same thing using rewriting and L'Hopital's (with the minus sign).2011-04-27
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    @Arturo: Aha, I had made a silly mistake. I placed a negative sign in front because I had thought (mistakenly) that the limit clearly went to infinity. I now agree with you and Carl.2011-04-27
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    @Carl, that's the correct answer. When you say you rewrote the arctangent, how did you rewrite it into something that goes to zero?2011-04-27
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    nevermind, got it2011-04-27

2 Answers 2

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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, $$\lim_{n\to\infty}\left(\frac{1}{2}+\frac{1}{\pi}\arctan\frac{nx}{t}\right)^n = \mathrm{exp}\left(-\frac{t}{x\pi}\right).$$

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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 ${\displaystyle e^{-{t \over \pi x}}}$