How to find the limit of this equation? I have a Calc final coming up soon and am currently reviewing.  I am currently stumped on finding the limit of as $x \rightarrow 1$, of $(1-\sqrt{2x^2 -1}) / (x-1)$.
Don't you have to multiply by the conjugate?  Any help is seriously appreciated.  I know the answer is $-2$, but I cannot seem to get that.

Here's a way to compute the limit without using derivatives but by multiplying by the conjugate:
\begin{align*}
\lim_{x\to1} \frac{1 - \sqrt{2x^2-1}}{x-1} \cdot \frac{1 + \sqrt{2x^2-1}}{1 + \sqrt{2x^2-1}}
&= \lim_{x\to1} \frac{1 - (2x^2-1)}{(x-1)(1 + \sqrt{2x^2-1})} \\
&= \lim_{x\to1} \frac{-2x^2 + 2}{(x-1)(1 + \sqrt{2x^2-1})} \\
&= \lim_{x\to1} \frac{-2(x^2 - 1)}{(x-1)(1 + \sqrt{2x^2-1})} \\
&= \lim_{x\to1} \frac{-2(x - 1)(x + 1)}{(x-1)(1 + \sqrt{2x^2-1})} \\
&= \lim_{x\to1} \frac{-2(x + 1)}{1 + \sqrt{2x^2-1}} \\
&= \frac{-2(1 + 1)}{1 + \sqrt{2(1)^2-1}} \\
&= -2
\end{align*}
