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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.
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September 2nd, 2017, 07:38 AM   #2
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Quote:
Originally Posted by deltaX View Post
Tex doesn't seem to work.
Use [math]...[/math]. I've edited your post.
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September 2nd, 2017, 09:32 AM   #3
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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 $\delta = \min{(\frac \epsilon M, \frac12)}$.
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Last edited by v8archie; September 2nd, 2017 at 09:47 AM.
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September 2nd, 2017, 06:25 PM   #4
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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.
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September 2nd, 2017, 09:07 PM   #5
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Here's a useful reference for/explanation of such proofs: Epsilon Delta Proofs
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September 3rd, 2017, 03:47 AM   #6
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Quote:
Originally Posted by deltaX View Post
$\displaystyle \frac{1}{2} < \left|x\right|< \frac{3}{2}$ 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, $\left|\frac{(1)^3 + 1}{(1)^3}\right| = 2 > \frac{35}{27}$

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.
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Last edited by cjem; September 3rd, 2017 at 04:01 AM.
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September 3rd, 2017, 06:24 AM   #7
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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.)
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September 3rd, 2017, 09:49 AM   #8
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* absolute value of the expression.
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September 3rd, 2017, 05:05 PM   #9
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Quote:
Originally Posted by cjem View Post
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 $\delta$.
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September 4th, 2017, 06:43 AM   #10
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Quote:
Originally Posted by v8archie View Post
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 $|x - a| \lt \delta \implies |f(x)-L| \lt \epsilon$
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.
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Last edited by cjem; September 4th, 2017 at 06:53 AM.
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