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February 16th, 2017, 01:37 AM   #1
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Basic limit problem

So I kind of know what to do, but I'm stuck in the process of finding a delta.

lim(x^(2)+4x)
x-->-4

I started with this:
* |(x^(2)+4x)-0|<(epsilon)

Then
* |(x^(2)+4x)|<(epsilon)

Then I factored:
* |x(x+4)|<(epsilon)

I know I'm trying to get "|x+4|<(epsilon)," which I could then set to "<(delta)," But I can't divide to get rid of the x on the left, because I can't have an x on the right either. This is as far as I got, that is, if everything is correct so far.
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February 16th, 2017, 04:15 AM   #2
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Quote:
Originally Posted by nbg273 View Post
So I kind of know what to do, but I'm stuck in the process of finding a delta.

lim(x^(2)+4x)
x-->-4

I started with this:
* |(x^(2)+4x)-0|<(epsilon)

Then
* |(x^(2)+4x)|<(epsilon)

Then I factored:
* |x(x+4)|<(epsilon)

I know I'm trying to get "|x+4|<(epsilon)," which I could then set to "<(delta)," But I can't divide to get rid of the x on the left, because I can't have an x on the right either. This is as far as I got, that is, if everything is correct so far.
One would expect that the limiting function is equal to the function value at the point, so we would expect that

$\displaystyle \begin{align*} \lim_{x \to -4} \left( x^2 + 4\,x \right) &= \left( -4 \right) ^2 + 4\,\left( -4 \right) \\ &= 16 - 16 \\ &= 0 \end{align*}$

To prove it, we have to show that for all $\displaystyle \begin{align*} \epsilon > 0 \end{align*}$ there exists a $\displaystyle \begin{align*} \delta \end{align*}$ such that $\displaystyle \begin{align*} 0 < \left| x - \left( -4 \right) \right| < \delta \implies \left| x^2 + 4\,x - 0 \right| < \epsilon \end{align*}$, so

$\displaystyle \begin{align*} \left| x^2 + 4\,x \right| &< \epsilon \\ \left| x \right| \left| x + 4 \right| &< \epsilon \\ \left| x + 4 - 4 \right| \left| x + 4 \right| &< \epsilon \\ \left( \left| x + 4 \right| - 4 \right) \left| x + 4 \right| \leq \left| x + 4 - 4 \right| \left| x + 4 \right| &< \epsilon \textrm{ by the Reverse Triangle Inequality} \\ \left( \left| x + 4 \right| - 4 \right) \left| x + 4 \right| &< \epsilon \\ \left| x + 4 \right| ^2 - 4\,\left| x + 4 \right| &< \epsilon \\ \left| x + 4 \right| ^2 - 4 \,\left| x + 4 \right| + \left( -2 \right) ^2 &< \epsilon + \left( -2 \right) ^2 \\ \left( \left| x + 4 \right| - 2 \right) ^2 &< \epsilon + 4 \\ \left| \left| x + 4 \right| - 2 \right| &< \sqrt{ \epsilon + 4 } \\ -\sqrt{ \epsilon + 4 } < \left| x + 4 \right| - 2 &< \sqrt{ \epsilon + 4} \\ 2 - \sqrt{\epsilon + 4} < \left| x + 4 \right| &< 2 + \sqrt{ \epsilon + 4} \end{align*}$

So we can let $\displaystyle \begin{align*} \delta = 2 + \sqrt{ \epsilon + 4 } \end{align*}$ and by reversing every step we can show that the implication holds, and thus the limit is 0.
Thanks from topsquark

Last edited by skipjack; February 16th, 2017 at 07:12 AM.
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February 16th, 2017, 04:21 AM   #3
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If $|x-4| \lt \delta$ then
$$\begin{gather*}-\delta \lt x-4 \lt \delta \\
4 - \delta \lt x \lt 4+\delta \end{gather*}$$

So, if we say that we will always take $\delta \le 1$ we have
$$3 \lt x \lt 5$$

Can you continue?
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February 16th, 2017, 04:29 AM   #4
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Quote:
Originally Posted by v8archie View Post
If $|x-4| \lt \delta$ then
$$\begin{gather*}-\delta \lt x-4 \lt \delta \\
4 - \delta \lt x \lt 4+\delta \end{gather*}$$

So, if we say that we will always take $\delta \le 1$ we have
$$3 \lt x \lt 5$$

Can you continue?
It's not $\displaystyle \begin{align*} \left| x - 4 \right| \end{align*}$, it's $\displaystyle \begin{align*} \left| x - \left( -4 \right) \right| \end{align*}$...
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February 16th, 2017, 09:49 AM   #5
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Quote:
Originally Posted by Prove It View Post
So we can let $\delta = 2 + \sqrt{ \epsilon + 4 }$ . . .
Did you mean "So we can let $\delta = -2 + \sqrt{ \epsilon + 4 }$ . . ."?
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February 16th, 2017, 10:08 AM   #6
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Quote:
Originally Posted by Prove It View Post
It's not $\displaystyle \begin{align*} \left| x - 4 \right| \end{align*}$, it's $\displaystyle \begin{align*} \left| x - \left( -4 \right) \right| \end{align*}$...
So it is, but the principal holds anyway.
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February 18th, 2017, 01:44 AM   #7
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Quote:
Originally Posted by skipjack View Post
Did you mean "So we can let $\delta = -2 + \sqrt{ \epsilon + 4 }$ . . ."?
Since when do you undo a minus 2 by subtracting 2 from both sides?
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February 18th, 2017, 02:48 AM   #8
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Since when would $\delta$ not be small when $\epsilon$ is small?
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