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September 2nd, 2017, 07:13 AM   #1
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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: $\displaystyle |f(x) - L| = |\frac{(x+1)^2(x^2-x+1)}{x^3}|$. Not sure where to go from here.

EDIT: Tex doesn't seem to work.
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Last edited by greg1313; September 2nd, 2017 at 07:37 AM.

September 2nd, 2017, 07:38 AM   #2
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Quote:
 Originally Posted by deltaX Tex doesn't seem to work.
Use $...$. I've edited your post.

 September 2nd, 2017, 09:32 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,091 Thanks: 2360 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 $\delta = \min{(\frac \epsilon M, \frac12)}$. Thanks from greg1313 and deltaX 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 $\displaystyle \frac{1}{2} < \left|x\right|< \frac{3}{2}$ 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: 7,091 Thanks: 2360 Math Focus: Mainly analysis and algebra Here's a useful reference for/explanation of such proofs: Epsilon Delta Proofs Thanks from jonah and greg1313
September 3rd, 2017, 03:47 AM   #6
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Quote:
 Originally Posted by deltaX $\displaystyle \frac{1}{2} < \left|x\right|< \frac{3}{2}$ implies $\displaystyle \left|\frac{x^3+1}{x^3}\right| < \frac{(\frac{3}{2})^3+1}{(\frac{3}{2})^3} = \frac{\frac{35}{8}}{\frac{27}{8}} = \frac{35}{27}$
This is not correct. For example, $\left|\frac{(1)^3 + 1}{(1)^3}\right| = 2 > \frac{35}{27}$

$\displaystyle \frac{1}{2} < \left|x\right|< \frac{3}{2}$ 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
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Quote:
 Originally Posted by deltaX Thanks for the helpful answer. Here is my next attempt to make the proof. Please comment if I'm doing something wrong.
To make life easier, you can write $$\frac{x^3+1}{x^3}=1+\frac1{x^3}$$
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: 7,091 Thanks: 2360 Math Focus: Mainly analysis and algebra * absolute value of the expression.
September 3rd, 2017, 05:05 PM   #9
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Quote:
 Originally Posted by cjem If you instead say $\displaystyle \frac{1}{2} < \left|x\right|< \frac{3}{2}$ 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.
Note that
$$-\frac{3}{2} \lt x \lt -\frac{1}{2} \implies \left|\frac{x^3+1}{x^3}\right| \lt \left| \frac{(-\frac{1}{2})^3+1}{(-\frac{1}{2})^3}\right| = \frac{-\frac{7}{8}}{-\frac{1}{8}} = 7$$
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|\left|1+\frac1{x^3}\right|\end{align*}
Also, $$\begin{array}{c c c}|x + 1| \lt \delta \le \frac12 & \implies & -\tfrac12 \lt x+1 \lt \tfrac12 \\ && -\tfrac32 \lt x \lt -\tfrac12 \\ && -\tfrac{27}8 \lt x^3 \lt -\tfrac18 \\ && -8 \lt \tfrac1{x^3} \lt -\tfrac8{27} \\ && -7 \lt 1 + \tfrac1{x^3} \lt \tfrac{19}{27} \end{array}$$
And thus $$\left|1+\frac1{x^3}\right| \lt 7$$
We also have $|x+1| \lt \delta \le \frac{\epsilon}{7}$ and so \begin{align*} \left|\frac{x^4 + x + 1}{x^3} + 1\right| &= |x+1|\left|1+\frac1{x^3}\right| \\ &\lt \frac{\epsilon}{7} \cdot 7 = \epsilon \\ &&\blacksquare\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
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Quote:
 Originally Posted by v8archie The definition is that the limit $\displaystyle \lim_{x \to a} f(x) = L$ ifgiven any $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that $|x - a| \lt \delta \implies |f(x)-L| \lt \epsilon$
Being nitpicky, this is stronger than the definition of a limit. It should read "... such that $0 < |x - a| < \delta \implies$ ..." rather than "... such that $|x-a| < \delta \implies$ ..."

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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