My Math Forum Basic limit problem

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 February 16th, 2017, 01:37 AM #1 Member   Joined: Feb 2017 From: East U.S. Posts: 33 Thanks: 0 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.
February 16th, 2017, 04:15 AM   #2
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Quote:
 Originally Posted by nbg273 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.

Last edited by skipjack; February 16th, 2017 at 07:12 AM.

 February 16th, 2017, 04:21 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,784 Thanks: 2197 Math Focus: Mainly analysis and algebra 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?
February 16th, 2017, 04:29 AM   #4
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Quote:
 Originally Posted by v8archie 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*}...

February 16th, 2017, 09:49 AM   #5
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Quote:
 Originally Posted by Prove It So we can let $\delta = 2 + \sqrt{ \epsilon + 4 }$ . . .
Did you mean "So we can let $\delta = -2 + \sqrt{ \epsilon + 4 }$ . . ."?

February 16th, 2017, 10:08 AM   #6
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Quote:
 Originally Posted by Prove It 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.

February 18th, 2017, 01:44 AM   #7
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Quote:
 Originally Posted by skipjack 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?

 February 18th, 2017, 02:48 AM #8 Global Moderator   Joined: Dec 2006 Posts: 17,221 Thanks: 1294 Since when would $\delta$ not be small when $\epsilon$ is small?

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