Prove a function $f(x)$ has a limit as $x$ approaches $1$. I am trying to prove the function $f(x) = (1-x)/(1-\sqrt{x})$ has a limit as $x$ approaches 1 using an epsilon definition. 
I've gotten as far as finding the limit is $2$ by factoring the numerator, as well as setting up the proof, which is as follows:
Let $\epsilon>0$ be given. Then there exists $\delta>0$ such that $|(1-x)/(1-\sqrt{x}) - 2| < \epsilon$ if $0 < x-1 < \delta$, $x$ element of the domain, and $1$ is an accumulation point.
Now, simplifying the absolute value gives $\left|\frac{(1-x) - 2(1-\sqrt{x})}{1-\sqrt{x}}\right| < \epsilon$, but now I do not know how to proceed.
 A: hint:$\left|\dfrac{1-x}{1-\sqrt{x}}-2\right| = \left|\dfrac{-1+2\sqrt{x}-x}{1-\sqrt{x}}\right|=\dfrac{(1-\sqrt{x})^2}{|1-\sqrt{x}|}=|1-\sqrt{x}|= \dfrac{|1-x|}{1+\sqrt{x}}\leq |1-x|$, and the $\epsilon-\delta$ argument can be used.
A: For $x \neq 1$, note that
$$\dfrac{1-x}{1-\sqrt{x}} = \dfrac{1-(\sqrt{x})^2}{1-\sqrt{x}} = \dfrac{(1+\sqrt{x})(1-\sqrt{x})}{1-\sqrt{x}} = 1+\sqrt{x}$$
Now this enable you in simplifying your limit expression.
A: If you want to use the usual definition, you have to find the limit first! If $x = 1 + \tau$ for some small $\tau$, then $\sqrt{x} \approx 1+\tau/2$. So
$$
\frac{1-x}{1-\sqrt{x}} \approx \frac{-\tau}{-\tau/2} = 2.
$$
So the conjecture is that your function has the limit $2$ as $x$ tends to $1$.
At this point, it is a good idea to subtract your function from $2$:
$$
2 - \frac{1-x}{1-\sqrt{x}} = \frac{1-2\sqrt{x}+x}{1-\sqrt{x}} = \frac{(1-\sqrt{x})^2}{1-\sqrt{x}} = 1-\sqrt{x}.
$$
As $x$ tends to $1$, $\sqrt{x}$ also tends to $1$, and so the limit is $2$.
You take it from here.
A: Hint
$$\frac{1 - x}{1 - \sqrt{x}} = \frac{1 - x}{1 - \sqrt{x}} \cdot \frac{1 + \sqrt{x}}{1 + \sqrt{x}}.$$
