Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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.

share|cite|improve this question
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
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 $x=1$.

share|cite|improve this answer
Thank you, kind sir! – UnfoLimited Mar 9 '12 at 10:49

1+√1=2; however √1−1=0

2/0 -> infinity This simple...

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.