<s>[INST] Finding the limit I need to find the limit of this problem. I pretty much know you have to multiply by the conjugate but I get lost after I do that.
$$\lim\limits_{x\to 1} \frac{(1 / \sqrt{x}) - 1}{1-x}$$
 [/INST] $$\begin{eqnarray}
-\lim\limits_{x\to 1} \frac{1 / \sqrt{x} - 1}{1-x} &=& \lim\limits_{x\to 1} \frac{1 / \sqrt{x} - 1}{x-1}\\
&=& \lim\limits_{x\to 1} \frac{1 / \sqrt{x} - 1}{x-1}\frac{1/\sqrt{x}+1}{1/\sqrt{x}+1}\\
&=& \lim\limits_{x\to 1} \frac{1/x - 1}{(x-1)(1/\sqrt{x}+1)}\\
&=& \lim\limits_{x\to 1} \frac{1}{1/\sqrt{x}+1}\times \lim\limits_{x\to 1}\frac{1/x - 1}{x-1}\\
&=& \frac{1}{2}\times \lim\limits_{x\to 1}\frac{1/x - 1}{x-1}
\end{eqnarray}$$
This limit can be evaluated by noting that substituting $1/x$ for $x$ gives one over the limit, but should give the same value since $x\to 1$ is the same as $1/x\to 1$, thus the limit $L$ satisfies $L=1/L$, so $L=1$. This gives us a final answer of $-1/2$. 
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