# 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?

• Hint: try the substitution $t=\sqrt{x}$ from which: 1. $t\to 1$ for $x\to 1$ and $x=t^2$. ;) – Ixion Sep 8 '18 at 23:35
• since $x\to 1$ you can assume $x>0$ and use $1-x = (1-\sqrt{x})(1+\sqrt{x})$ – Alan Muniz Sep 8 '18 at 23:37
• You could also try Bernoulli's Rule. $d/dx (1-\sqrt(x))=1/(2\sqrt{x})$ and $d/dx (1-x)=-1$. $1/(2\sqrt{1})=1/2$. – Shrey Joshi Sep 9 '18 at 1:05
• $$\lim_{x\to1}\frac{1-\sqrt{x}}{1-x}=\left\{\frac{0}{0}\right\}=\lim_{x\to1}\frac{\left[1-\sqrt{x}\right]'}{[1-x]'}=\lim_{x\to1}\frac{1}{2\sqrt{x}}=1$$ – Hazem Orabi Sep 9 '18 at 7:16
• @AlanMuniz Feel free to write an answer, I'd happily accept that. – Max Sep 11 '18 at 0:04

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 $$\lim_{x \to 1} \frac{1 - \sqrt x}{1 - x} = \lim_{x \to 1} \frac{1 - \sqrt x}{(1 - \sqrt x)(1 + \sqrt x)} = \lim_{x \to 1} \frac{1}{1 + \sqrt x} = \frac{1}{2}$$
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}
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 …
• So we multiply both sides by $-1$ to change the signs. Then $\lim_{x \to 1} \sqrt x = 1$, and $\lim_{x \to 1} 1 = 1$ and $\lim_{x \to 1} x-1 = 0$. Isn't that the same problem as above, that I have $0$ in the denominator? I'm slow here, but I can't seem to see what was the reason for changing the signs? – Max Sep 9 '18 at 7:43
• Oh! just to make evident the rate of variation of $f$. – Bernard Sep 9 '18 at 8:57