# Thread: Using epsilon-delta to prove lim{x>-2}(x^4)=16

1. ## Using epsilon-delta to prove lim{x>-2}(x^4)=16

Hi all,

I've tried a lot of times, but I'm still stuck on $\lim_{x\to-2}(x^4)=16$

So far, I've done:

To show that $0< |x+2|< \delta$ exists for all $|x^4-16|< \epsilon$
$
|x^4-16|$

$=|(x^2+4)(x-2)(x+2)|$

$= ...$

$=|(x+2)^2 - 4(x+2) + 8||x+2-4||x+2|$

However, I can't use the normal triangle inequality here since there's a negative number inside.

I've also tried:
$|x^4-16|< \epsilon$
$-\epsilon + 16 < x^4 < \epsilon + 16$

$\sqrt[4]{-\epsilon + 16} < x < \sqrt[4]{\epsilon + 16}$
So the limit exists since $x=-2$ always exists in that interval for all $\epsilon > 0$.

But that is not the epsilon delta proof I need.

Thanks.

2. Originally Posted by owq
Hi all,

I've tried a lot of times, but I'm still stuck on $\lim_{x\to-2}(x^4)=16$
If $0 < |x+2| < \delta$ then $- \delta < x+2 < \delta$ so $- 2 - \delta < x < -2 + \delta$. If you let $\delta \leq 2$ then $- 2 - \delta < x \leq 0$. This means $|x| < |-2-\delta| = \delta + 2$. Now you need to make $|x^4 - 16|$ arbitrary small. Notice that $|x^4 - 16| = |x-2||x^3+2x^2 + 4x + 8|$. Now apply triangle inequality, $|x^4-16| \leq |x-2| (|x|^3 +2|x|^2 + 4|x| + 8)$. Apply the fact that $|x-2| < \delta$ and $|x| < \delta + 2 \leq 4$. Therefore, $|x^4-16| < \delta (64 + 32 + 16 + 8) = 120\delta$. If you want to make $|x^4-16| < \epsilon$, set $\delta = \min \{ 2, \tfrac{\epsilon}{120} \}$.

3. ## Burning questions

Thanks for the explanation! I have some questions to ask, if you don't mind.

Originally Posted by ThePerfectHacker
If you let $\delta \leq 2$ then $- 2 - \delta < x \leq 0$. This means $|x| < |-2-\delta| = \delta + 2$.
But if $|x| < \delta + 2$, then wouldn't it be $-\delta - 2 < x < \delta + 2$? Which would exceed the boundaries of $- 2 - \delta < x \leq 0$?

Originally Posted by ThePerfectHacker
Apply the fact that $|x-2| < \delta$ and $|x| < \delta + 2 \leq 4$.
Wasn't $|x+2| < \delta$?

Thanks.

4. Here is another way.
$\left| {x + 2} \right| < 1\, \Rightarrow \,\left| x \right| < 3\, \Rightarrow \,\left| {x - 2} \right| < 5\;\& \,\left| {x^2 + 2} \right| < 11$.

Suppose that $\varepsilon > 0$ then choose $\delta = \min \left\{ {1,\frac{\varepsilon }{{55}}} \right\}$.

If $\left| {x + 2} \right| < \delta$ then $\left| {x^4 - 16} \right| = \left| {x + 2} \right|\left| {x - 2} \right|\left| {x^2 + 2} \right| < \delta \cdot 5 \cdot 11 < \varepsilon$.

5. Originally Posted by owq
But if $|x| < \delta + 2$, then wouldn't it be $-\delta - 2 < x < \delta + 2$? Which would exceed the boundaries of $- 2 - \delta < x \leq 0$?[/tex]
No. We showed if $0<|x+2| < \delta$ then $|x|<\delta + 2$. This is a true statement, you can look at the above proof or draw the intervals on the number line. What you are saying is that if $|x| < \delta + 2$ then $-\delta - 2 but you need to remember that when $|x| < \delta + 2$ it does can contain all the values of $x$ it contains those values of $x$ such that $0 < |x+2| < \delta$ so it never gets out of its interval.

Wasn't $|x+2| < \delta$?
Sorry, I was not careful there.
You can write, $|x^4-16| = |x+2||x-2||x^2-4| \leq |x+2|(|x|+2)(|x|^2+4)$ where $|x|\leq 4$ and $|x+2|<\delta$ to get your result now.

6. ## I see

Thank you both!

Originally Posted by ThePerfectHacker
You can write, $|x^4-16| = |x+2||x-2||x^2-4| \leq |x+2|(|x|+2)(|x|^2+4)$ where $|x|\leq 4$ and $|x+2|<\delta$ to get your result now.
We can apply the triangle inequality like this? I thought that we would have to use the lower bounds triangle inequality here since there's a difference of two numbers for $|x-2|$ and $|x^2-4|$

7. Originally Posted by owq
Thank you both!

We can apply the triangle inequality like this? I thought that we would have to use the lower bounds triangle inequality here since there's a difference of two numbers for $|x-2|$ and $|x^2-4|$
$|x-2| = |x+(-2)| \leq |x| + |-2| = |x|+|2|$.

There is also, $|x-2|\geq ||x|-|2||$ but we are not using it here.

8. ## Boundaries

Ok, I was having the wrong concept of the triangle inequality. Thanks for clearing that up! Can't believe I overlooked it! Now I'm still confused about the boundaries of $0 < |x-a| < \delta$

Originally Posted by Plato
$\left| {x + 2} \right| < 1\, \Rightarrow \,\left| x \right| < 3\, \Rightarrow \,\left| {x - 2} \right| < 5\;\& \,\left| {x^2 + 2} \right| < 11$.
Originally Posted by ThePerfectHacker
No. We showed if $0<|x+2| < \delta$ then $|x|<\delta + 2$. This is a true statement, you can look at the above proof or draw the intervals on the number line. What you are saying is that if $|x| < \delta + 2$ then $-\delta - 2 but you need to remember that when $|x| < \delta + 2$ it does can contain all the values of $x$ it contains those values of $x$ such that $0 < |x+2| < \delta$ so it never gets out of its interval.
Ok, I see that when $|x| < \delta + 2$ it contains all the values from $0<|x+2| < \delta$.

But aren't we trying to fulfill the inequality $0<|x+2| < \delta$? If $|x| < \delta + 2$ then let's say that when $x = \delta + 1.9$, it will fulfill $|x| < \delta + 2$ but not $0<|x+2| < \delta$ which was our original intention.

Similarly for $\left|{x + 2} \right| < 1\, \Rightarrow \,\left| x \right| < 3\$, let's say $x = 2$, then it would fulfill $|x|< 3$ but not $|x + 2| < 1$.

Did I get my concepts wrong somewhere?

9. Originally Posted by owq
Ok, I see that when $|x| < \delta + 2$ it contains all the values from $0<|x+2| < \delta$.

But aren't we trying to fulfill the inequality $0<|x+2| < \delta$? If $|x| < \delta + 2$ then let's say that when $x = \delta + 1.9$, it will fulfill $|x| < \delta + 2$ but not $0<|x+2| < \delta$ which was our original intention.

Similarly for $\left|{x + 2} \right| < 1\, \Rightarrow \,\left| x \right| < 3\$, let's say $x = 2$, then it would fulfill $|x|< 3$ but not $|x + 2| < 1$.

Did I get my concepts wrong somewhere?
If $0<|x+2|<\delta$ then $|x|<\delta + 2$ (for $0<\delta \leq 2$).

That is all we proved. You are saying but some values in $|x| <\delta + 2$ do not satisfy $0<|x+2|<\delta$. So what? You are thinking that if $|x|<\delta + 2$ then $0<|x+2|<\delta$. This is clearly not true, just take $x=-2$. When we restrain $0<|x+2|<\delta$ all the inequalities used in the proof becomes satisfied. But this does not mean that any other additional numbers which satisfy the other inequalities must satisfy our first inequality that we started with. We do not have to worry about that because we have $x$ constrained.