My Math Forum Basic limit problem
 User Name Remember Me? Password

 Calculus Calculus Math Forum

 February 16th, 2017, 12:37 AM #1 Member   Joined: Feb 2017 From: East U.S. Posts: 40 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, 03:15 AM   #2
Member

Joined: Oct 2016
From: Melbourne

Posts: 77
Thanks: 35

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 06:12 AM.

 February 16th, 2017, 03:21 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,214 Thanks: 2410 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, 03:29 AM   #4
Member

Joined: Oct 2016
From: Melbourne

Posts: 77
Thanks: 35

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, 08:49 AM   #5
Global Moderator

Joined: Dec 2006

Posts: 18,683
Thanks: 1522

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, 09:08 AM   #6
Math Team

Joined: Dec 2013
From: Colombia

Posts: 7,214
Thanks: 2410

Math Focus: Mainly analysis and algebra
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, 12:44 AM   #7
Member

Joined: Oct 2016
From: Melbourne

Posts: 77
Thanks: 35

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, 01:48 AM #8 Global Moderator   Joined: Dec 2006 Posts: 18,683 Thanks: 1522 Since when would $\delta$ not be small when $\epsilon$ is small?

 Tags basic, limit, problem

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post SammyG Elementary Math 3 November 14th, 2015 07:20 PM DerGiLLster Calculus 1 September 16th, 2015 06:44 PM Tutu Calculus 4 June 26th, 2012 04:08 AM sjeddie Calculus 3 August 28th, 2008 07:45 AM balibalo Advanced Statistics 2 February 11th, 2008 04:37 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top