# Alternative ways to find $\lim_{x\to 1}\; \frac{1-x}{1- \sqrt{x}}$

Searching on Google by a solution, I found this:

$1) \space$The numerator can be written by a diference of squares:$\space (1-\sqrt{x})(1+\sqrt{x})$

$2 \space)$Then, one can "eliminated" the commum factor beteween the numerator and the denominator.

$3) \space$The final expression looks like: $$\lim_{x\to 1}\; {1+ \sqrt{x}}=2$$

However this was not a intuitive algebraic solution. I haven't thought of this solution at the first attempts. Could you please give me other alternative algebraic solutions, if it exists. Thanks.

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Same solution, let $x=u^2$, but now instantly familiar. – André Nicolas Jan 7 '13 at 17:20
Why isn't it intuitive? If you have a $0/0$ form, the first thing one usually does is search for common factors. – David Mitra Jan 7 '13 at 17:21
The difference of two squares method was the first thing that popped into my head before I clicked the link to this question! – Clive Newstead Jan 7 '13 at 17:21
The other approach to this solution is to think - that was neat, how will I be able to spot neat solutions like that next time and the comment from @AndréNicolas was exactly what I was going to put. – Mark Bennet Jan 7 '13 at 17:22
@João What's $(1-\sqrt{x})(1+\sqrt{x})$? – WimC Jan 7 '13 at 17:35

You can compute the inverse: $$\lim_{x\to 1}\frac{1-\sqrt{x}}{1-x}=\lim_{x\to 1}\frac{\sqrt{x}-1}{x-1}$$ which is, by definition, $f'(1)$ where $f(x)=\sqrt{x}$.

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As I understood, you "turn over" the fraction,by puting $\frac{1-\sqrt{x}}{1-x}$ on the denominator. Then you multiplied the denominator by $\frac{-1}{-1}$ to get the final expression, that is under the main fraction (the numerator is $1$).Finally you computed the derivative of $\sqrt{x}$ at $x=1$ to get $\frac{1}{\frac{1}{2}}$. Is that? – João Jan 7 '13 at 18:04
Yeah, that's basically it. Essentially, if $g(x)/h(x)\to A$ then $h(x)/g(x)\to 1/A$ – Thomas Andrews Jan 8 '13 at 14:12

Same solution, let $x=u^2$, but now instantly familiar.

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Perhaps we can use L'Hospital rule here ?

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By substitution:$\quad$ let $x = u^2\,.\quad$Then as $x\to 1,\;\;u^2\to 1\implies u\to 1$, giving us $$\lim_{x\to 1}\; \frac{1-x}{1- \sqrt{x}} \;\;=\;\; \lim_{u\to 1}\; \frac{1-u^2}{1- u} = \lim_{x\to 1} \frac{u^2 - 1}{u-1} \;=\;\lim_{u \to 1} \frac{(u-1)(u+1)}{u-1} = \lim_{u\to 1}(u + 1) = 2$$

Using the substitution makes the "difference of squares" route looks so much more obvious!

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If $u^2 \to 1$, then $u \to 1$ or $u \to -1$. Is this second situation if one compute the limit we get $0$.M'I rigth? – João Jan 7 '13 at 18:14
since $x \to 1$, the only root of concern is u = 1 (since $u = \sqrt{x}$ must necessarily be $\ge 0$ in this situation. – amWhy Jan 7 '13 at 18:22