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$\lim_{x\to1}\frac{x + \sqrt{x}}{\sqrt{x-1}}$ $\lim_{x\to1}\frac{x - \sqrt{x}}{\sqrt{x-1}}$

Lately I've been trying to satisfy some curiosity about the nature of limits and I found this example, it's really bugging me that I can't solve it. It seems so simple, yet when I attempt to solve it, it's like I'm trapped in a loop. Everything I do, undos the previous action.

I'll take anything you can give, a solution, a good hint... I tried manipulating the expression without changing it, but I can't get through it.

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The numerator goes to $2$ while the denominator goes to $0$. The limit is infinity. Did you mean $x\color{Red}-\sqrt{x}$, in which case the limit goes to $0$? –  anon Mar 9 '12 at 10:21
For x=(+1), it's undefined. You may try to take the square of the equation I think... –  Kerim Atasoy Mar 9 '12 at 10:22
Wait, that's the conclusion? That simple? I thought that was it, I just kept pushing and pushing to get something explicit. Nifty. No, thanks, now it's all clear. :D –  UnfoLimited Mar 9 '12 at 10:24
How would it go if it were $x - \sqrt{x}$? –  UnfoLimited Mar 9 '12 at 10:29

2 Answers 2

up vote 3 down vote accepted

The numerator tends toward $2$ while the denominator tends toward $0$; the form $2/0$ is not indeterminate at all, so no energy needs be spent beating it into shape when we can see for a fact just from this that the limit is $\infty$ (i.e. for any $N>0$ as large as you want there is a neighborhood of the argument $x=1$ for which the ratio is larger than $N$ throughout the entire neighborhood).

In the alternate situation, with a minus sign we can multiply/divide by the numerator's conjugate,

$$\frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x-1}}\cdot \frac{\sqrt{x}+1}{\sqrt{x}+1}=\frac{\sqrt{x}\sqrt{x-1}}{\sqrt{x}+1}.$$

This allows for us to directly plug in $x=1$.

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Thank you, kind sir! –  UnfoLimited Mar 9 '12 at 10:49

1+√1=2; however √1−1=0

2/0 -> infinity This simple...

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