limit problem

**jayshizwiz** · December 1st 2010, 10:31 PM

Prove $\displaystyle \lim_{x\to 1}\frac{x+2}{x^2+2}=1$ using $\: \epsilon, \: \delta$

Attempt:

$\: \epsilon >0$ .
We must find $\delta >0$ so that every $x$ that fulfills $0<|x-1|< \delta \:\:\:$ fulfills $|\frac{x+2}{x^2+2} -1| < \epsilon$

$|\frac{x+2}{x^2+2}-1| = |\frac{x-x^2}{x^2+2}| = \frac{|x||1-x|}{|x^2+2|} = \frac{|x||x-1|}{|x^2+2|}$

I'm not sure about this part:

$\frac{|x||x-1|}{|x^2+2|} \leq_{\downarrow} \frac{|x||x-1|}{|x|} = |x-1| < \epsilon$
since $_{|x^2+2| \geq |x|}$

and since $0<|x-1|< \delta$ ,

for every $0 < \delta \leq \epsilon$ the required is fulfilled.

Is this correct?
Thanks!

**matheagle** · December 1st 2010, 11:14 PM

How about the same idea with the squeeze theorem?

$0<\left|{x+2\over x^2+2}-1\right|=\left|{x-x^2\over x^2+2}\right|={|x||1-x|\over x^2+2}< |1-x|\to 0$

**jayshizwiz** · December 2nd 2010, 01:22 AM

you can use the squeeze theorem, but the question specifically says use epsilon delta proof.

But does it look right the way I did it?

**imiviortal** · December 2nd 2010, 01:43 AM

yes i almost sure that you are right .

**HallsofIvy** · December 2nd 2010, 02:15 AM

No longer necessary.

**imiviortal** · December 2nd 2010, 02:41 AM

yes i know ive edited my message one minute before youve posted hehe

(3:14)

**jayshizwiz** · December 2nd 2010, 10:06 PM

I didn't see what you guys originally wrote but i'm going to assume it was funny.

**HallsofIvy** · December 3rd 2010, 04:28 AM

Not funny- imiviortal had a typo. I pointed it out but he had corrected it while I was typing so I removed my post.

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