Prove that for $\displaystyle f:\mathbb{R}\to\mathbb{R}$ defined as $\displaystyle f(x)=2x^2+8$, $\displaystyle f$ has a limit at $\displaystyle 2$.

Obviously the limit is $\displaystyle 16$. So we need to show that for each $\displaystyle \epsilon>0$, there exists $\displaystyle \delta>0$ such that for each $\displaystyle x$ satisfying:

$\displaystyle |x-2|<\delta$

it is then true that:

$\displaystyle |(2x^2+8)-16|<\epsilon$

So let's get busy. The last bit simplifies to:

$\displaystyle |x^2-4|<\frac{\epsilon}{2}$

I can only guess what to do next. I could try this:

$\displaystyle -\frac{\epsilon}{2}<x^2-4<\frac{\epsilon}{2}$

which simplifies to:

$\displaystyle 4-\frac{\epsilon}{2}<x^2<4+\frac{\epsilon}{2}$

But this is a little strange. I want to take the square root, but the left side could be negative, so I can't. So, maybe I could do:

$\displaystyle max\{0,4-\frac{\epsilon}{2}\}<x^2<4+\frac{\epsilon}{2}$

which simplifies to:

$\displaystyle \sqrt{max\{0,4-\frac{\epsilon}{2}\}}<|x|<\sqrt{4+\frac{\epsilon}{ 2}}$

But how would I get that into something workable?

Alternatively, we could go back to this:

$\displaystyle -\frac{\epsilon}{2}<x^2-4<\frac{\epsilon}{2}$

And try this:

$\displaystyle -\frac{\epsilon}{2}<(x-2)(x+2)<\frac{\epsilon}{2}$

But then we have a problem of negatives. So let's assume $\displaystyle x+2<0$. Then:

$\displaystyle -\frac{\epsilon}{2(x+2)}>(x-2)>\frac{\epsilon}{2(x+2)}$

$\displaystyle |x-2|<-\frac{\epsilon}{2(x+2)}$

Now assume $\displaystyle x+2\geq0$. Then:

$\displaystyle |x-2|<\frac{\epsilon}{2(x+2)}$

In both cases we have this:

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

But still we have a problem. How do we find $\displaystyle \delta$ from this:

$\displaystyle \delta\leq\frac{\epsilon}{2|x+2|}$

?

Or am I going about this entirely the wrong way?

Any help would be much appreciated!