Evaluating a limit of a Fourier series I have a Fourier series representation of a solution to the heat equation, given by
$\displaystyle u(t,x) = \\ \displaystyle\sin \omega t + \sum_{n = 1}^{\infty}{\frac{4( - 1)^{n}}{(2n - 1)\pi}\omega\left\lbrack \frac{\omega\sin \omega t + \left( n - \frac{1}{2} \right)^{2}\pi^{2}\left( \cos \omega t - e^{- \left( n - \frac{1}{2} \right)^{2}\pi^{2}t} \right)}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}} \right\rbrack\cos\left\lbrack \left( n - \frac{1}{2} \right)\pi x \right\rbrack}.$
How would one go about computing
$\displaystyle \lim_{\omega\rightarrow\infty}{u(t,0)}$
explicitly?
 A: A quicker method to obtain the same result is as follows. Firstly, merge the $\sin\omega t$ term with the rest of the Fourier series. This is achieved by noting that the coefficient of $\sin\omega t$ is unity, and so the Fourier series of unity in terms of the spatial variable $x$ may be employed:
\begin{align*}
\sin\omega t &= \sum_{n=1}^{\infty}{\frac{4\left( -1 \right)^{n+1}}{(2n-1)\pi}\sin\omega t\cos\left[\left( n - \frac{1}{2}\right)\pi x\right]} \\
&= - \sum_{n = 1}^{\infty}{\frac{4( - 1)^{n}}{(2n - 1)\pi}\sin\omega t\cos\left\lbrack \left( n - \frac{1}{2} \right)\pi x \right\rbrack}.
\end{align*}
Thus
\begin{align*}
u(t,x) &= \sum_{n = 1}^{\infty}{\frac{4( - 1)^{n}}{(2n - 1)\pi}\omega\left\lbrack \frac{\omega\sin\omega t + \left( n - \frac{1}{2} \right)^{2}\pi^{2}\left( \cos\omega t - e^{- \left( n - \frac{1}{2} \right)^{2}\pi^{2}t} \right)}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}} - \frac{\sin\omega t}{\omega} \right\rbrack\cos\left\lbrack \left( n - \frac{1}{2} \right)\pi x \right\rbrack}.
\end{align*}
Hence,
\begin{align*}
u(t,0) &= \sum_{n = 1}^{\infty}{\frac{4( - 1)^{n}}{(2n - 1)\pi}\left\lbrack \frac{\omega^{2}\sin\omega t + \left( n - \frac{1}{2} \right)^{2}\pi^{2}\omega\left( \cos\omega t - e^{- \left( n - \frac{1}{2} \right)^{2}\pi^{2}t} \right)}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}} - \sin\omega t \right\rbrack} \\
&= \sum_{n = 1}^{\infty}{\frac{4( - 1)^{n}}{(2n - 1)\pi}\frac{\omega^{2}\sin\omega t + \left( n - \frac{1}{2} \right)^{2}\pi^{2}\omega\left( \cos\omega t - e^{- \left( n - \frac{1}{2} \right)^{2}\pi^{2}t} \right) - \left( \omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4} \right)\sin\omega t}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}}} \\
&= \sum_{n = 1}^{\infty}{\frac{4( - 1)^{n}}{(2n - 1)\pi}\frac{\left( n - \frac{1}{2} \right)^{2}\pi^{2}\omega\left( \cos\omega t - e^{- \left( n - \frac{1}{2} \right)^{2}\pi^{2}t} \right) - \left( n - \frac{1}{2} \right)^{4}\pi^{4}\sin\omega t}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}}}.
\end{align*}
To determine the limit as $\omega \rightarrow \infty$ of $u(t,0)$, consider each term separately:
\begin{align*}
\text{Limit 1:} && \lim_{\omega \rightarrow \infty}\left\lbrack \frac{\left( n - \frac{1}{2} \right)^{2}\pi^{2}\omega\left( \cos\omega t \right)}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}} \right\rbrack, \\
\text{Limit 2:} && \lim_{\omega \rightarrow \infty}\left\lbrack \frac{\left( n - \frac{1}{2} \right)^{2}\pi^{2}\omega\left( - e^{- \left( n - \frac{1}{2} \right)^{2}\pi^{2}t} \right)}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}} \right\rbrack, \\
\text{Limit 3:} && \lim_{\omega \rightarrow \infty}\left\lbrack \frac{- \left( n - \frac{1}{2} \right)^{4}\pi^{4}\sin\omega t}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}} \right\rbrack,
\end{align*}
which all tend to zero, giving
\begin{equation*}
\lim_{\omega\rightarrow\infty}{u(t,0)} = 0
\end{equation*}
as required.
