February 16th, 2017, 01:37 AM  #1 
Newbie Joined: Feb 2017 From: East U.S. Posts: 14 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  
Member Joined: Oct 2016 From: Melbourne Posts: 68 Thanks: 33  Quote:
$\displaystyle \begin{align*} \lim_{x \to 4} \left( x^2 + 4\,x \right) &= \left( 4 \right) ^2 + 4\,\left( 4 \right) \\ &= 16  16 \\ &= 0 \end{align*}$ 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: 6,399 Thanks: 2105 Math Focus: Mainly analysis and algebra 
If $x4 \lt \delta$ then $$\begin{gather*}\delta \lt x4 \lt \delta \\ 4  \delta \lt x \lt 4+\delta \end{gather*}$$ 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 
Member Joined: Oct 2016 From: Melbourne Posts: 68 Thanks: 33  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 
Global Moderator Joined: Dec 2006 Posts: 16,401 Thanks: 1177  
February 16th, 2017, 10:08 AM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,399 Thanks: 2105 Math Focus: Mainly analysis and algebra  
February 18th, 2017, 01:44 AM  #7 
Member Joined: Oct 2016 From: Melbourne Posts: 68 Thanks: 33  
February 18th, 2017, 02:48 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 16,401 Thanks: 1177 
Since when would $\delta$ not be small when $\epsilon$ is small?


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