
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 $\displaystyle 0<x1< \delta \:\:\:$ fulfills $\displaystyle \frac{x+2}{x^2+2} 1 < \epsilon$
$\displaystyle \frac{x+2}{x^2+2}1 = \frac{xx^2}{x^2+2} = \frac{x1x}{x^2+2} = \frac{xx1}{x^2+2}$
I'm not sure about this part:
$\displaystyle \frac{xx1}{x^2+2} \leq_{\downarrow} \frac{xx1}{x} = x1 < \epsilon$
since $\displaystyle _{x^2+2 \geq x}$
and since $\displaystyle 0<x1< \delta$,
for every $\displaystyle 0 < \delta \leq \epsilon$ the required is fulfilled.
Is this correct?
Thanks!

How about the same idea with the squeeze theorem?
$\displaystyle 0<\left{x+2\over x^2+2}1\right=\left{xx^2\over x^2+2}\right={x1x\over x^2+2}< 1x\to 0$

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?

yes i almost sure that you are right .


yes i know ive edited my message one minute before youve posted hehe :) (3:14)

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

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