Lateral limits of function involving hyperbolic trignometric functions

Question

I am not being able to calculate the lateral limits at 0 of the following function

$f(x) = \frac{\sinh(x)}{2\sqrt{\cosh(x) - 1}}$

I have tried both direct substitution (yields 0/0) and L'Hospital's rule (yields (0/0)/0 )

Answer 1

Write $\sinh x$ as $\frac{e^x-e^{-x}}2$ and $\cosh x$ as $\frac{e^x+e^{-x}}2$, then substitute $t=e^x$. $$\begin{align}\lim_{x\to0}\frac{\sinh x}{2\sqrt{\cosh x-1}}&=\lim_{x\to0}\frac{\frac{e^{x}-e^{-x}}2}{2\sqrt{\frac{e^x+e^{-x}}2-1}}\\&=\lim_{t\to1}\frac{\frac{t^2-1}{2t}}{2\sqrt{\frac{t^2-2t+1}{2t}}}\\&=\lim_{t\to1}\frac{(t+1)(t-1)\sqrt{2t}}{(t-1)4t}\\&=\lim_{t\to1}\frac{(t+1)\sqrt{2t}}{4t}\\&=\frac{\sqrt2}{2}\end{align}$$ EDIT

I forget to put absolute value brackets to $t-1$ after canceling square root, so two-sided limit doesn't exist. For $x\to0^-$ limit will be $-\frac{\sqrt{2}}2$ and for $x\to0^+$ limit will be $\frac{\sqrt{2}}2$.

Answer 2

I can suggest for you this way of finding this limit, using the Taylor series expansion of $\sinh(x)$ and $\cosh(x)$ near zero, to the order one . we have $$\lim_{x\rightarrow 0} \frac{\sinh(x)}{2\sqrt{\cosh(x)-1}} = \lim_{x\rightarrow 0} \frac{x}{2\sqrt{1+ x^2/2-1}} =\frac{1}{\sqrt{2}} $$

Answer 3

Since $\cosh(2x) =2\sinh^2(2x)+1 $ and $\sinh(2x) =2\sinh(x)\cosh(x) $,

$\begin{align*} f(x) &= \frac{\sinh(x)}{2\sqrt{\cosh(x) - 1}}\\ &= \frac{\sinh(x)}{2\sqrt{2\sinh^2(x/2)}}\\ &= \frac{\sinh(x)}{2\sqrt{2}\sinh(x/2)}\\ &= \frac{2\sinh(x/2)\cosh(x/2)}{2\sqrt{2}\sinh(x/2)}\\ &= \frac{\cosh(x/2)}{\sqrt{2}}\\ &\to \frac{1}{\sqrt{2}} \qquad\text{since } \cosh(x) \to 1 \text{ as }x \to 0\\ \end{align*} $