# [SOLVED] Prove the limit of this polynomial

• Mar 18th 2009, 04:06 PM
hatsoff
[SOLVED] Prove the limit of this polynomial
Prove that for $f:\mathbb{R}\to\mathbb{R}$ defined as $f(x)=2x^2+8$, $f$ has a limit at $2$.

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

$|x-2|<\delta$

it is then true that:

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

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

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

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

$-\frac{\epsilon}{2}

which simplifies to:

$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:

$max\{0,4-\frac{\epsilon}{2}\}

which simplifies to:

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

But how would I get that into something workable?

Alternatively, we could go back to this:

$-\frac{\epsilon}{2}

And try this:

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

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

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

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

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

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

In both cases we have this:

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

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

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

?

Any help would be much appreciated!
• Mar 18th 2009, 05:57 PM
Reckoner
Quote:

Originally Posted by hatsoff
$|x^2-4|<\frac{\epsilon}{2}$

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

Let's do

$\lvert x^2-4\rvert<\frac\epsilon2$

$\Rightarrow\lvert(x-2)(x+2)\rvert<\frac\epsilon2$

$\Rightarrow\lvert x-2\rvert\lvert x+2\rvert<\frac\epsilon2$

Now, for $0<\lvert x-2\rvert<1$ (i.e., if we only consider $\delta\leq1$), we have $3 and

$\lvert x-2\rvert\lvert x+2\rvert<\frac\epsilon2$

$\Leftarrow5\lvert x-2\rvert<\frac\epsilon2$

$\Leftrightarrow\lvert x-2\rvert<\frac\epsilon{10}$

So we can choose

$\delta=\min\left\{1,\frac\epsilon{10}\right\}.$

Now, working through backwards, you can write a clean proof.
• Mar 18th 2009, 06:20 PM
Plato
Here is a way I like to start these. Suppose that $\left| {x - 2} \right| < 1$ then $\left| {x + 2} \right| < 3$.
Thus $\left| {f(x) - f(2)} \right| = \left| {2x^2 - 8} \right| = 2\left| {x - 2} \right|\left| {x + 2}\right| \leqslant 6\left| {x - 2} \right|$.
Now if $\varepsilon > 0$ then pick $\delta = \min \left[ {1,\frac{\varepsilon }{6}} \right]$.
• Mar 18th 2009, 07:41 PM
hatsoff
Thanks a million, guys! It's so simple... I must be getting old.