# Homework Help: Quick question on a limit

1. Jul 13, 2010

### SimpliciusH

I was copying some old text and I came across a limit I didn't understand.

It starts as

$$\stackrel{Lim}{x\rightarrow0}\frac{\sqrt{x^2+2-1}-x}{x}$$

and then understandably continues until

$$\stackrel{Lim}{x\rightarrow0}\frac{x-1}{x*\sqrt{x^2+x-1}+x}=0$$

Why would this be zero? x-1 goes to -1, x goes to zero and anything multiplied by zero is zero. And dividing with zero is a no no...

Sorry for the bad format I'm still trying to get a hang of latex.

Last edited: Jul 13, 2010
2. Jul 13, 2010

### ehild

The second limit has no sense. The expression under the square root becomes negative when x--->0.

ehild

3. Jul 13, 2010

### SimpliciusH

The second limit is derived from the first one. I know the second one makes no sense but its quite confidently written that it equals zero.

Is the zero perhaps a reference to the first limit? And there was a mistake or typo made during solving?

4. Jul 13, 2010

You must have made at least one transcription error.

$$\lim_{x \to 0} \frac{\sqrt{x^2+2-1}-x}{x}$$

does not exist - the expression goes to infinity IF what you have beneath the square root is correct.

The second expression

$$\lim_{x \to 0} \frac{x-1}{x\,\sqrt{x^2 + x -1} +x}$$

does not equal zero - it too goes to infinity (note that the denominator is

$$x \left(\sqrt{x^2+x-1} + 1\right)$$

and this goes to zero as $x$ itself does. More importantly, this does not come from your first expression.