# limit problem

• Dec 1st 2010, 10:31 PM
jayshizwiz
limit problem
Prove $\displaystyle \displaystyle \lim_{x\to 1}\frac{x+2}{x^2+2}=1$ using $\displaystyle \: \epsilon, \: \delta$

Attempt:

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

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

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

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

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

Is this correct?
Thanks!
• Dec 1st 2010, 11:14 PM
matheagle
How about the same idea with the squeeze theorem?

$\displaystyle 0<\left|{x+2\over x^2+2}-1\right|=\left|{x-x^2\over x^2+2}\right|={|x||1-x|\over x^2+2}< |1-x|\to 0$
• Dec 2nd 2010, 01:22 AM
jayshizwiz
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?
• Dec 2nd 2010, 01:43 AM
imiviortal
yes i almost sure that you are right .
• Dec 2nd 2010, 02:15 AM
HallsofIvy
No longer necessary.
• Dec 2nd 2010, 02:41 AM
imiviortal
yes i know ive edited my message one minute before youve posted hehe :) (3:14)
• Dec 2nd 2010, 10:06 PM
jayshizwiz
I didn't see what you guys originally wrote but i'm going to assume it was funny.
• Dec 3rd 2010, 04:28 AM
HallsofIvy
Not funny- imiviortal had a typo. I pointed it out but he had corrected it while I was typing so I removed my post.