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.
$$\lim\limits_{x\to 1} \frac{(1 / \sqrt{x}) - 1}{1-x}$$
 A: You don't have to multiply by a conjugate. Hint: $1-x=(1-\sqrt{x})(1+\sqrt{x})$.
A: You need to find
$$\lim\limits_{x \to 1}\frac{\frac 1 {\sqrt x}-1}{1-x}$$
This is 
$$\mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt x }}\frac{{1 - \sqrt x }}{{1-x}}$$
Can you move on with that? (With anon's hint maybe?)
A: \begin{align}
\lim_{x \to 1} \dfrac{1/\sqrt{x}-1}{1-x} & = \lim_{x \to 1} \dfrac1{\sqrt{x}}\dfrac{1-\sqrt{x}}{1 - (\sqrt{x})^2} = \lim_{x \to 1} \dfrac1{\sqrt{x}}\dfrac{1-\sqrt{x}}{(1-\sqrt{x})(1+\sqrt{x})}\\
& = \lim_{x \to 1} \dfrac1{\sqrt{x}}\dfrac1{(\sqrt{x}+1)} = \dfrac12
\end{align}
A: $$\begin{eqnarray}
-\lim\limits_{x\to 1} \frac{1 / \sqrt{x} - 1}{1-x} &=& \lim\limits_{x\to 1} \frac{1 / \sqrt{x} - 1}{x-1}\\
&=& \lim\limits_{x\to 1} \frac{1 / \sqrt{x} - 1}{x-1}\frac{1/\sqrt{x}+1}{1/\sqrt{x}+1}\\
&=& \lim\limits_{x\to 1} \frac{1/x - 1}{(x-1)(1/\sqrt{x}+1)}\\
&=& \lim\limits_{x\to 1} \frac{1}{1/\sqrt{x}+1}\times \lim\limits_{x\to 1}\frac{1/x - 1}{x-1}\\
&=& \frac{1}{2}\times \lim\limits_{x\to 1}\frac{1/x - 1}{x-1}
\end{eqnarray}$$
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 $-1/2$. 
A: I'd opt for L'Hôpital's rule.
\begin{align}
&L = \lim_{x \to 1} \dfrac{\dfrac{1}{\sqrt{x}}-1}{1-x}\\
&L = \lim_{x \to 1} \dfrac{1}{-2x^{\frac{3}{2}}} \cdot -1\\
&L = \lim_{x \to 1} \dfrac{1}{2x^\frac{3}{2}}\\
&L = \dfrac{1}{2}\\
\end{align}
A: Let's solve it elementarily:
$$\lim\limits_{x\to 1} \frac{(1 / \sqrt{x}) - 1}{1-x}=\lim\limits_{x\to 1} \frac{(1 / \sqrt{x}) - 1}{1-x} \cdot \frac{(1 / \sqrt{x}) + 1}{(1 / \sqrt{x}) + 1}=\lim\limits_{x\to 1}\frac{1-x}{2x(1-x)}=\frac{1}{2}.$$
Q.E.D.
A: $\lim\limits_{x\to 1} \frac{(1 / \sqrt{x}) - 1}{1-x}$.
We substition $\sqrt {x}=t$, hance we:
If $ x\longrightarrow 1\Rightarrow t\longrightarrow 1. $
From here for the given limits have:
$\lim\limits_{x\to 1} \frac{(1 / \sqrt{x}) - 1}{1-x}$=$\lim\limits_{t\to 1} \frac{\frac{1}{t} - 1}{1-t^2}$=$\lim\limits_{t\to 1} \frac{\frac{1-t}{t}}{1-t^2}$=$\lim\limits_{t\to 1} \frac{1-t}{t(1-t)(1+t)}$=$\lim\limits_{t\to 1} \frac{1}{t(t+1)}$=$\frac{1}{2}$
