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Math Help - Using epsilon-delta to prove lim{x>-2}(x^4)=16

  1. #1
    owq
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    Exclamation 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
    <br />
|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.

    Kindly advise. Need a hint or something. I'm brain dead. Lol.

    Thanks.
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  2. #2
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    Quote Originally Posted by owq View Post
    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} \}.
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  3. #3
    owq
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    Burning questions

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

    Quote Originally Posted by ThePerfectHacker View Post
    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?

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

    Thanks.
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    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 .
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    Quote Originally Posted by owq View Post
    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<x<\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.
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  6. #6
    owq
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    I see

    Thank you both!

    Quote Originally Posted by ThePerfectHacker View Post
    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|
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    Quote Originally Posted by owq View Post
    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.
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  8. #8
    owq
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    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

    Quote Originally Posted by Plato View Post
    \left| {x + 2} \right| < 1\, \Rightarrow \,\left| x \right| < 3\, \Rightarrow \,\left| {x - 2} \right| < 5\;\& \,\left| {x^2  + 2} \right| < 11.
    Quote Originally Posted by ThePerfectHacker View Post
    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<x<\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?
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    Quote Originally Posted by owq View Post
    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.
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