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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?

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

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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}

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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 …

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  • $\begingroup$ 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? $\endgroup$ – Max Sep 9 '18 at 7:43
  • $\begingroup$ Oh! just to make evident the rate of variation of $f$. $\endgroup$ – Bernard Sep 9 '18 at 8:57

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