How to apply limit properties in this case? I was asked to find the following limit:
$$
\lim_{x \to 1} \frac{1 - \sqrt x}{1 - x}
$$
I worked it out using direct substitution that the limit is $\frac{1}{2}$.
Initially I was trying a more algebraic approach, finding separately the limits of 
$$
\lim_{x \to 1} 1 - \sqrt x = 0
$$
$$
\lim_{x \to 1} 1 - x = 0
$$
And then applying limit properties for division and multiplication: 
$$
\lim_{x \to a} (f \cdot g)(x) = l \cdot m
$$
$$
\lim_{x \to a} (\frac{1}{g})(x) = \frac{1}{m}
$$
But that doesn't work, since the limit in the denominator will $0$. 
So besides direct substitution, the limit properties are of no use in this case when there is a $0$ in the denominator?
 A: If it were not for cases like this, where it is NOT possible to find the limit just by finding the limits of the numerator and denominator separately, one would probably never use the concept of limits at all.
The answer by Alan Muniz gives you one way to do this. Here's another:
\begin{align}
& \lim_{x\,\to\,1} \frac{1-\sqrt x}{1-x} = \lim_{u\,\to\,\text{what?}} \frac{1-u}{1-u^2} \\[10pt]
& \text{(As $x\to1,$ then $u = \sqrt x \to \sqrt 1 = 1$.)} \\[10pt]
= {} & \lim_{u\,\to\,1} \frac{1-u}{(1-u)(1+u)} = \lim_{u\,\to\,1} \frac 1 {1+u} = \cdots
\end{align}
A: Hint;
Change the signs of the numerator and denominator to get
$$\frac{\sqrt x-1}{x-1}=\frac{f(x)-f(1)}{x-1}, \enspace\text{where } f(x)=\sqrt x$$
This is a rate of variation from $x=1$, hence the limit is …
A: Since the function involves a square root of $x$ we tacitly assume $x >0$ and from $x\to 1$ we assume that $x\neq 1$. Then we may use $1- x = (1 - \sqrt x)(1 + \sqrt x)$. It follows that
$$
\lim_{x \to 1} \frac{1 - \sqrt x}{1 - x} = \lim_{x \to 1} \frac{1 - \sqrt x}{(1 - \sqrt x)(1 + \sqrt x)} = \lim_{x \to 1} \frac{1}{1 + \sqrt x} = \frac{1}{2}
$$
