Why is the limit at x=0 equal to zero in this calculus problem?

[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.

 

[ehild]

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

ehild

 

[SimpliciusH]

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

ehild

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?

 

[statdad]

You must have made at least one transcription error.

Your first expression

<br /> \lim_{x \to 0} \frac{\sqrt{x^2+2-1}-x}{x}<br />

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

The second expression

<br /> \lim_{x \to 0} \frac{x-1}{x\,\sqrt{x^2 + x -1} +x}<br />

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

<br /> x \left(\sqrt{x^2+x-1} + 1\right)<br />

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

Please examine your original problem and repost.

I could attempt to "guess" different forms for the correct expression, but:
- there is no guarantee would ever hit the correct one, even though I'm reasonably sure I would)
- the weather is fantastic, my bicycle is ready to go, and there is a 55-mile ride mapped out that has my name on it. hoo-rah!

 

[SimpliciusH]

Thanks for taking a look at this. I've found the same problem solved on some other notes I was doing and found the transcript error.