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

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

I'm not sure about this part:

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

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!