Calculate a limit: $\lim_{t \to 0+} {1\over 2}\cdot({\pi \over t}\cdot\frac{1+e^{-2\pi t}}{1 - e^{-2\pi t}} - {1\over t^2}) $ Please calculate the limit $\lim\limits_{t \to 0+} {1\over 2}\cdot \left({\pi \over t}\cdot\frac{1+e^{-2\pi t}}{1 - e^{-2\pi t}} - {1\over t^2}\right)$ and provide the corresponding procedure.
The answer is $\pi^2 \over 6$.
I tried L'Hospital's Rule but I failed.
The following is not necessary to read:

The background of this problem:
this problem arose from a problem related to the Fourier Transform. I tried to use Poisson's summation formula to get $\sum\limits_{n = -\infty}^\infty \frac{1}{n^2 + t^2} = {\pi \over t}\cdot\frac{1+e^{-2\pi t}}{1 - e^{-2\pi t}}$ . (This can be proved by let $f(x) = \frac{1}{x^2+t^2}$, where $t>0$ is a parameter. Then $\hat{f}(n) = {\pi \over t}\cdot e^{-2\pi |n| t}$.) Then $\sum\limits_{n = 1}^\infty {1\over n^2} = \frac{\pi^2}{6} $ should be a corollary.

 A: By L'Hospital:
With $u=\pi t$, the expression is
$$\frac{\pi^2}2\left(\frac1{u\tanh u}-\frac1{u^2}\right)=\frac{\pi^2}2\left(\frac{u-\tanh u}{u^2\tanh u}\right).$$
As $\dfrac{\tanh u}u$ tends to one, we can replace the denominator by $u^3$. Then
$$\lim_{u\to0}\frac{u-\tanh u}{u^3}=\lim_{u\to0}\frac{1-1+\tanh^2u}{3u^2}=\frac13.$$
Finally,
$$\frac{\pi^2}6.$$
A: Not very detailed, hope everyone can understand.
$\begin{align*}
  \sum\limits_{n = 1}^\infty {1\over n^2}= &\lim\limits_{t \to 0+} {1\over 2}\cdot \left({\pi \over t}\cdot\frac{1+e^{-2\pi t}}{1 - e^{-2\pi t}} - {1\over t^2}\right) \\
  = & {1\over 2}\cdot \lim\limits_{t \to 0+} \left({\pi \over t}\cdot\frac{1+e^{-2\pi t}}{1 - e^{-2\pi t}} - {1\over t^2}\right)\\
  = & {1\over 2}\cdot \lim\limits_{t \to 0+} \frac{\pi t (1 + e^{-2\pi t} )-1 + e^{-2\pi t}}{t^2(1 - e^{-2\pi t})}\\
  = & {1\over 2}\cdot \lim\limits_{t \to 0+}\frac{\pi t + \pi t e^{-2\pi t}-1 + e^{-2\pi t}}{2\pi t^3} \\
  = & {1\over 2}\cdot \lim\limits_{t \to 0+}\frac{\pi + \pi e^{-2\pi t} - 2\pi^2 e^{-2\pi t} - 2 \pi e^{-2\pi t}}{6\pi t^2}\\
  = & {1\over 2}\cdot \lim\limits_{t \to 0+} \frac{4 \pi^3 t e^{-2 \pi t}}{12 \pi t} \\
  = & {1\over 2}\cdot \lim\limits_{t \to 0+} \frac{4\pi^3 e^{-2\pi t}-8\pi^4 t e^{-2\pi t}}{12 \pi}\\
  = & {1\over 2} \cdot {4\pi^3 \over12 \pi}\\
  = &{\pi^2 \over 6}
  \end{align*}$
A: Use Taylor series of an exponential function so that a direct computation shows 
\begin{align*}&\frac{1}{2} \bigg(\frac{\pi}{t}\frac{1+e^{-2\pi t} }{1-e^{-2pi t} } -
\frac{1}{t^2}  \bigg)
\\&= \frac{1}{2t}  \bigg\{ \pi \frac{2+(-2\pi t) + \frac{(-2\pi t)^2}{2}+\cdots }{
(-2\pi t) + \frac{(-2\pi t)^2}{2}+\cdots }(-1) - \frac{1}{t} \bigg\}
\\&= \frac{1}{4 t^2 }  \bigg\{  \frac{
2+(-2\pi t) + \frac{(-2\pi t)^2}{2}+\cdots }{ 1 + \frac{(-2\pi
t)^1}{2} +  \frac{(-2\pi t)^2}{3!}+\cdots }-2 \bigg\}
\\&= \frac{1}{4t^2}
 \frac{ \frac{(-2\pi t)^2}{6}+\cdots }{ 1+ \frac{(-2\pi t)}{2}+\cdots }\end{align*}
A: Expanding @labbhattarcharjee's comment, we use $\coth x\approx\frac{1}{x}+\frac{x}{3}$ to write the limit as $$\lim_{t\to0^+}\frac12\left(\frac{\pi}{t}\coth(\pi t)-\frac{1}{t^2}\right)\lim_{t\to0^+}=\frac12\left(\frac{\pi}{t}\frac{1}{\pi t}+\frac{\pi}{t}\frac{\pi t}{3}-\frac{1}{t^2}\right)=\frac{\pi^2}{6}$$(the first term cancels the last one).
