We start with the definition of the limit:
$$\forall \epsilon , \exists \delta \text{ s.t.}$$
$$\left\vert f(x)-L \right\vert \lt\epsilon,$$
$$0\lt\left\vert x-a \right\vert \lt\delta$$
So we begin with some preliminary steps before writing out our proof:
$$\left\vert\frac{x-4}{x^2+2} + \frac{1}{11}\right\vert\lt\epsilon\Rightarrow$$
$$\frac{1}{11}\left\vert\frac{11x-44+x^2+2}{x^2+2}\right\vert\lt\epsilon\Rightarrow$$
$$\frac{1}{11}\left\vert x-3\right\vert\left\vert x+14\right\vert\left\vert\frac{1}{x^2+2}\right\vert\lt\epsilon\Rightarrow$$
Now I proceed to bound the $\left\vert x+14\right\vert$ and $\left\vert\frac{1}{x^2+2}\right\vert$ terms:
Say $\left\vert x-3\right\vert\lt2\Rightarrow$
$-2\lt x-3\lt 2$ $\space$ $\space$ which implies $\space$ $\space$ $15\lt x+14\lt19$
And here is where my problem arises. I don't know how to bound the $\left\vert\frac{1}{x^2+2}\right\vert$ term.
Any help would be appreciated, from hints to a well-established proof.
Best Answer
Why don't you try to use the fact that $$x^2\geqslant 0?$$