Limits of a function involving $\mathrm{cn}(x,k)$ Given $$f(x) = \frac{1 - \mathrm{cn}(x,k)}{{\sqrt3}(1+\mathrm{cn}(x,k)) - 1 + \mathrm{cn}(x,k)}$$
what would be $$\lim_{x\to 0} f(x)$$ 
and $$\lim_{x\to\infty} f(x)$$ when 
$$k=\frac{\sqrt{2-\sqrt{3}}}{2}?$$
 A: For $x=0$, use the Taylor series:
$$
\mathrm{cn}(x,k) = 1 - \frac{1}{2}   x^{2} + \Biggl(\frac{1}{24} + \frac{k^{2}}{6}\Biggr)   x^{4} - \Biggl(\frac{1}{720} + \frac{11 k^{2}}{180} + \frac{k^{4}}{45}\Biggr)   x^{6} + \operatorname{O} \bigl(x^{8}\bigr)
$$
to conclude that
$$
\lim_{x\to 0}\frac{1 - \mathrm{cn} (x,k)}{\sqrt{3}(1 + \mathrm{cn} (x,k)) - 1 + \mathrm{cn} (x,k)} =
\lim_{x\to0}\left[\frac{\sqrt{3}}{12}   x^{2} + \Biggl(\frac{\sqrt{3}}{72} + \frac{1}{48} - \frac{\sqrt{3} k^{2}}{36}\Biggr)   x^{4} + \operatorname{O} \bigl(x^{6}\bigr)\right] = 0
$$
But for $x \to \infty$ ... it looks like your expression has an essential singularity at $x=\infty$ since your denominator has a sequence of zeros that go to infinity.
A: Here's a L'Hôpital-free way of doing things:
We can rearrange the expression a bit:
$$f(x)=\left({\sqrt3}\frac{1+\mathrm{cn}(x,k)}{1 - \mathrm{cn}(x,k)} - 1\right)^{-1}$$
use this,
$$f(x)=\left({\sqrt3}\mathrm{cd}^2\left(\frac{x}{2},k\right)\mathrm{ns}^2\left(\frac{x}{2},k\right) - 1\right)^{-1}$$
and transform back
$$f(x)=\frac{\mathrm{sn}^2\left(\frac{x}{2},k\right)}{\sqrt3 \mathrm{cd}^2\left(\frac{x}{2},k\right)- \mathrm{sn}^2\left(\frac{x}{2},k\right)}$$
and since $\mathrm{sn}(0,k)$ is $0$ while $\mathrm{cd}(0,k)$ is $1$, $f(0)=0$.

Here's another way to do
$$\lim_{x\to 0} \left({\sqrt3}\frac{1+\mathrm{cn}(x,k)}{1 - \mathrm{cn}(x,k)} - 1\right)^{-1}$$
Remember that $\mathrm{cn}(x,k)=\cos(\mathrm{am}(x,k))$, where $\mathrm{am}(x,k)$ (the Jacobian amplitude) is the inverse of the incomplete elliptic integral of the first kind; that is,
$$\phi=\mathrm{am}(x,k) \quad\leftrightarrow\quad x=F(\phi,k)$$
where
$$F(\phi,k)=\int_0^\phi\frac{\mathrm dt}{\sqrt{1-k^2\sin^2t}}$$
We can then determine that $\mathrm{am}(0,k)=0$, so we can perform an appropriate substitution and consider the limit
$$\lim_{t\to 0} \left({\sqrt3}\frac{1+\cos\,t}{1-\cos\,t}-1\right)^{-1}$$
thus,
$$\lim_{t\to 0} \left(\sqrt3\,\cot^2\frac{t}{2}-1\right)^{-1}=\lim_{t\to 0} \frac{\sin^2\frac{t}{2}}{\sqrt3\,\cos^2\frac{t}{2}-\sin^2\frac{t}{2}}=0$$
The procedure for the limit to infinity is similar (as $\phi\to\infty$, $F(\phi,k)\to\infty$ for $k^2 < 1$).
