September 2nd, 2017, 07:13 AM  #1 
Newbie Joined: Aug 2017 From: Norway Posts: 8 Thanks: 0  Epsilon delta proof
Hey all! So I have this problem I'm trying to solve. Use the epsilon delta definition of limits to prove that $\displaystyle \lim_{x\rightarrow (1)}\frac{x^4+x+1}{x^3}= 1 $ I've come up with this expression: $\displaystyle f(x)  L = \frac{(x+1)^2(x^2x+1)}{x^3}$. Not sure where to go from here. EDIT: Tex doesn't seem to work. Last edited by greg1313; September 2nd, 2017 at 07:37 AM. 
September 2nd, 2017, 07:38 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,641 Thanks: 959 Math Focus: Elementary mathematics and beyond  
September 2nd, 2017, 09:32 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,031 Thanks: 2342 Math Focus: Mainly analysis and algebra 
You don't need to take more than a linear factor out, so you can work with $$(x+1) \frac{x^3+1}{x^3}$$ The problem we have is that we need to limit the value of the quotient. This can be done by limiting the value of $x+1$. For example if $x+1 \lt \frac12$ (don't choose 1 because the quotient is unbounded at $x=0$). You can then determine the maximum absolute value $M$ (or any upper bound) of the quotient on $(\frac32,\frac12)$ and thus that $$\left(x+1) \frac{x^3+1}{x^3}\right \lt \frac12 M$$ for all $x+1 \lt \frac12$. Then, by limiting the size of $\delta$ to $\frac12$ you'll get a result. The proof will start by setting $\delta = \min{(\frac \epsilon M, \frac12)}$. Last edited by v8archie; September 2nd, 2017 at 09:47 AM. 
September 2nd, 2017, 06:25 PM  #4 
Newbie Joined: Aug 2017 From: Norway Posts: 8 Thanks: 0 
Thanks for the helpful answer. Here is my next attempt to make the proof. Please comment if I'm doing something wrong. Suppose that $\displaystyle \left x+1 \right < \frac{1}{2}$ implies $\displaystyle \frac{1}{2} < x+1< \frac{1}{2}$ implies $\displaystyle \frac{3}{2} < x< \frac{1}{2}$ implies $\displaystyle \frac{1}{2} < \leftx\right< \frac{3}{2}$ implies $\displaystyle \left\frac{x^3+1}{x^3}\right < \frac{(\frac{3}{2})^3+1}{(\frac{3}{2})^3} = \frac{\frac{35}{8}}{\frac{27}{8}} = \frac{35}{27}$ Then we have: $\displaystyle \left(x+1) \frac{x^3+1}{x^3} \right < \frac{35}{27} * \left x+1 \right.$ And $\displaystyle \left x+1 \right * \frac{35}{27} < \epsilon$ equivalent to $\displaystyle \left x+1 \right < \frac{\epsilon}{\frac{35}{27}} = \frac{27 \epsilon}{35}$ Given $\displaystyle \epsilon > 0$, we choose $\displaystyle \delta = min (\frac{1}{2},\frac{27\epsilon}{35}).$ Then we have that $\displaystyle 0 < \left x+1 \right < \delta = min (\frac{1}{2},\frac{27\epsilon}{35}).$ implies $\displaystyle \left(x+1) \frac{x^3+1}{x^3} \right < \frac {35}{27} * \frac{27\epsilon}{35} = \epsilon$ This completes the proof. 
September 2nd, 2017, 09:07 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,031 Thanks: 2342 Math Focus: Mainly analysis and algebra 
Here's a useful reference for/explanation of such proofs: Epsilon Delta Proofs 
September 3rd, 2017, 03:47 AM  #6  
Member Joined: Aug 2017 From: United Kingdom Posts: 97 Thanks: 28  Quote:
If you instead say $\displaystyle \frac{1}{2} < \leftx\right< \frac{3}{2}$ implies $\displaystyle \left\frac{x^3+1}{x^3}\right < \frac{(\frac{1}{2})^3+1}{(\frac{1}{2})^3} = \frac{\frac{9}{8}}{\frac{1}{8}} = 9$ and alter your subsequent work accordingly, you'll have a valid proof. Last edited by cjem; September 3rd, 2017 at 04:01 AM.  
September 3rd, 2017, 06:24 AM  #7  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,031 Thanks: 2342 Math Focus: Mainly analysis and algebra  Quote:
This makes for an easy calculation of an upper bound for the expression. (Be careful: $x$ is both small and negative.)  
September 3rd, 2017, 09:49 AM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,031 Thanks: 2342 Math Focus: Mainly analysis and algebra 
* absolute value of the expression.

September 3rd, 2017, 05:05 PM  #9  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,031 Thanks: 2342 Math Focus: Mainly analysis and algebra  Quote:
$$\frac{3}{2} \lt x \lt \frac{1}{2} \implies \left\frac{x^3+1}{x^3}\right \lt \left \frac{(\frac{1}{2})^3+1}{(\frac{1}{2})^3}\right = \frac{\frac{7}{8}}{\frac{1}{8}} = 7$$ but $M=9$ is perfectly acceptable, being an upper bound for the expression. It's just not the least upper bound. I think it would be more of a sketch of a proof  more detail should be provided to demonstrate that the $\delta\epsilon$ definition of the limit is satisfied. I'm going to use $M=7$ to demonstrate. The definition is that the limit $\displaystyle \lim_{x \to a} f(x) = L$ if given any $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that $$x  a \lt \delta \implies f(x)L \lt \epsilon$$In this case we claim that $\delta = \min{(\frac{\epsilon}{7},\frac12)}$ satisfies the definition. The proof of this can go as follows: We have \begin{align*}\left\frac{x^4 + x + 1}{x^3} + 1\right &= \left\frac{x^4 +x^3 + x + 1}{x^3}\right \\ &= x+1\left\frac{x^3 + 1}{x^3}\right = x+1\left1+\frac1{x^3}\right\end{align*}As you can see, the majority of the work is in determining $M$ and the required expression for $\delta$.  
September 4th, 2017, 06:43 AM  #10  
Member Joined: Aug 2017 From: United Kingdom Posts: 97 Thanks: 28  Quote:
Of course, in the case when $f$ is defined at $a$ and is continuous there (such as in this problem), your definition and the actual definition coincide. Last edited by cjem; September 4th, 2017 at 06:53 AM.  

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