<s>[INST] 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?
 [/INST] \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}
</s>