Find the limit of: $\lim_{x\to 1}\left(\frac{\sqrt{2x-1}-1}{x^2-1}\right)$ 
Find the limit of: $\lim\limits_{x\to 1}\left(\frac{\sqrt{2x-1}-1}{x^2-1}\right)$

I tried multiplying by the conjugate of $\sqrt{2x-1}-1$ ,but I obtain
$\frac{2x-2}{\left(x-1\right)\left(x+1\right)\left(\sqrt{2x-1}+1\right)}$,
which is again zero over zero indeterminate form.
How to approach this type of problems?

Another approach:
With the following substitution: $$\begin{aligned}\sqrt{2x-1}=t\implies x=&\ \frac{t^2+1}2\\\\\frac1{x+1}=&\ \frac2{t^2+3}\\\\\frac1{x-1}=&\ \frac2{t^2-1}=\frac2{(t-1)(t+1)}\ ,\end{aligned}$$
we have:
$$\lim_{x\to 1}\frac{\sqrt{2x-1}-1}{x^2-1}=\lim_{x\to 1}\frac{\sqrt{2x-1}-1}{(x-1)(x+1)}=\lim_{t\to 1}\left((t-1)\cdot\frac2{(t-1)(t+1)}\cdot\frac2{t^2+3}\right)=\frac12$$
