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?
 A: \begin{align}
L & = \lim_{x \to 1} \dfrac{\sqrt{2x-1}-1}{x^2-1} = \lim_{x \to 1} \dfrac{\sqrt{2x-1}-1}{x^2-1} \cdot \dfrac{\sqrt{2x-1}+1}{\sqrt{2x-1}+1}\\
& = \lim_{x \to 1} \dfrac{2x-1-1}{(x^2-1)(\sqrt{2x-1}+1)} = \lim_{x \to 1} \dfrac{2(x-1)}{(x^2-1)(\sqrt{2x-1}+1)}\\
& = \lim_{x \to 1} \dfrac2{(x+1)(\sqrt{2x-1}+1)} \left(\because \text{Canceling off }x-1\text{ from numerator and denominator}\right)\\
& = \dfrac2{2\cdot2}\\
& = \dfrac12
\end{align}
A: If you know L'Hôpital's rule,
\begin{align}
\lim_{x\to1}\frac{\sqrt{2x-1}-1}{x^2-1}&=\lim_{x\to1}\frac{\frac12(2x-1)^{-\frac12  }\cdot2}{2x}\\&=\lim_{x\to1}\frac1{2x}\cdot(2x-1)^{-\frac12}\\&=\frac12 
\end{align}
A: 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:
$$\lim_{x\to 1}\frac{\sqrt{2x-1}-1}{x^2-1}=\lim_{x\to 1}\frac{\sqrt{2x-1}-1}{(x-1)(x+1)}=\lim_{t\to 1}\left((t-1)\cdot\frac2{(t-1)(t+1)}\cdot\frac2{t^2+3}\right)=\frac12$$
A: By $t=x-1 \to 0$
$$\lim_{x\to 1}\frac{\sqrt{2x-1}-1}{x^2-1}=\lim_{t\to 0}\frac{\sqrt{2t+1}-1}{t(t+2)}$$
and by $f(t)=\sqrt{2t+1} \implies f'(0)=1$ we have
$$\lim_{t\to 0}\frac{\sqrt{2t+1}-1}{t(t+2)}=\lim_{t\to 0}\frac1{t+2}\cdot\frac{\sqrt{2t+1}-1}{t}=\frac12\cdot f'(0)=\frac12$$
