Show that $\lim_{a\to{0}}\frac{{\pi}a\ \coth{\mathrm{{\pi}a}-1}}{2a^2}=\frac{\pi^2}{6}$ How can I show that
$$\lim_{a\to{0}}\frac{{\pi}a\ \coth{\mathrm{{\pi}a}-1}}{2a^2}=\frac{\pi^2}{6}$$
I think the limit is in $\frac{0}{0}$ form, so I am using L'Hospital's rule, and then I cannot solve further, Please Help.
Thanks!
 A: Let $t = \pi a$. We then get that
$$\dfrac{\pi a \coth(\pi a)-1}{2a^2} = \pi^2 \dfrac{t \coth(t)-1}{2t^2}$$
Now let us look at $f(t) = \dfrac{t \coth(t)-1}{2t^2}$. We then have
$$f(t) = \dfrac{\coth(t)}{2t} - \dfrac1{2t^2}$$
Now use the Laurent series expansion. $$\coth(t) = \dfrac1t + \dfrac{t}3 + \mathcal{O}(t^3)$$
Hence, we have
$$f(t) = \dfrac{\dfrac1t + \dfrac{t}3 + \mathcal{O}(t^3)}{2t} - \dfrac1{2t^2} = \dfrac16 + \mathcal{O}(t^2)$$
Hence, we have
$$\lim_{t \to 0}f(t) =\dfrac16$$
Hence, we conclude that
$$\lim_{a \to 0}\dfrac{\pi a \coth(\pi a)-1}{2a^2} = \pi^2 \lim_{t \to 0}\dfrac{t \coth(t)-1}{2t^2} = \dfrac{\pi^2}6$$
A: Start with Marvis' approach, and evaluate $$\lim_{t\to{0}}\frac{t\ \coth(t)-1}{2t^2}=\lim_{t\to{0}}\frac{t\ \cosh(t)-\sinh(t)}{2t^2 \sinh(t)}$$
It is usually easier to work with sines and cosines. Apply l'Hopital rule once :
$$\lim_{t\to{0}}\frac{t\ \cosh(t)-\sinh(t)}{2t^2 \sinh(t)}=\lim_{t\to{0}}\frac{t\ \sinh(t)}{4 t \sinh(t) + 2t^2 \cosh(t)} $$
You can simplify $t$ in this last expression. Apply l'Hopital rule once more to find your result.
