What is the Epsilon-Delta Method for Proving Limits?

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In summary, the epsilon-delta method is used to show that the limit of a given function is equal to a specific value, in this case 3/2. By manipulating the expression and setting a restriction on the values of x, it can be proven that the limit is indeed 3/2.
  • #1
nycmathdad
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Use the epsilon-delta method to show that the limit is 3/2 for the given function.

lim (1 + 2x)/(3 - x) = 3/2
x-->1

I want to find a delta so that | x - 1| < delta implies |f(x) - L| < epsilon.

| (1 + 2x)/(3 - x) - (3/2) | < epsilon

-epsilon < (1 + 2x)/(3 - x) - 3/2 < epsilon

I now add 3/2 to all terms.

(3/2) - epsilon < (1 + 2x)/(3 - x) < (3/2) + epsilonStuck here...
 
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  • #2
You are nowhere near ready for $\displaystyle \epsilon \,\delta $ proofs, but for anyone playing at home...

$\displaystyle \begin{align*} \left| \frac{1 + 2\,x}{3 - x} - \frac{3}{2} \right| &< \epsilon \\
\left| \frac{2 \left( 1 + 2\,x \right) - 3 \left( 3 - x \right) }{ 2 \left( 3 - x \right) } \right| &< \epsilon \\
\left| \frac{2 + 4\,x - 9 + 3\,x}{2 \left( 3 - x \right) } \right| &< \epsilon \\
\left| \frac{7\,x - 7}{2 \left( 3 - x \right) } \right| &< \epsilon \\
\frac{7 \left| x - 1 \right| }{ 2 \left| 3 - x \right| } &< \epsilon \\
\frac{\left| x - 1 \right| }{\left| x - 3 \right| } &< \frac{2\,\epsilon}{7} \\
\left| x - 1 \right| &< \frac{2\,\epsilon}{7} \, \left| x - 3 \right| \end{align*} $

Now suppose we restrict $\displaystyle \left| x - 1 \right| < 1 $ for example, then

$\displaystyle \begin{align*} -1 < x - 1 &< 1 \\
-3 < x - 3 &< -1 \end{align*}$

so we can say for certain that if $\displaystyle \left| x - 1 \right| < 1 $, then $\displaystyle \left| x - 3 \right| < 3 $. Therefore

$\displaystyle \begin{align*} \left| x - 1 \right| < \frac{2\,\epsilon}{7}\,\left| x - 3 \right| &< \frac{2\,\epsilon}{7} \cdot 3 \\
\left| x - 1 \right| < \frac{2\,\epsilon}{7}\,\left| x - 3 \right| &< \frac{6\,\epsilon}{7} \\
\left| x - 1 \right| &< \frac{6\,\epsilon}{7} \end{align*} $

So we can finally get to the proof now...

Let $\displaystyle \delta = \min \left\{ 1, \frac{6\,\epsilon}{7} \right\} $, then reverse every step and you are done.
 
  • #3
Beer soaked ramblings follow.
Prove It said:
You are nowhere near ready for $\displaystyle \epsilon \,\delta $ proofs, but for anyone playing at home...
I think you just captured the essence of him who's perpetually stuck.
 
  • #4
Prove It said:
You are nowhere near ready for $\displaystyle \epsilon \,\delta $ proofs, but for anyone playing at home...

$\displaystyle \begin{align*} \left| \frac{1 + 2\,x}{3 - x} - \frac{3}{2} \right| &< \epsilon \\
\left| \frac{2 \left( 1 + 2\,x \right) - 3 \left( 3 - x \right) }{ 2 \left( 3 - x \right) } \right| &< \epsilon \\
\left| \frac{2 + 4\,x - 9 + 3\,x}{2 \left( 3 - x \right) } \right| &< \epsilon \\
\left| \frac{7\,x - 7}{2 \left( 3 - x \right) } \right| &< \epsilon \\
\frac{7 \left| x - 1 \right| }{ 2 \left| 3 - x \right| } &< \epsilon \\
\frac{\left| x - 1 \right| }{\left| x - 3 \right| } &< \frac{2\,\epsilon}{7} \\
\left| x - 1 \right| &< \frac{2\,\epsilon}{7} \, \left| x - 3 \right| \end{align*} $

Now suppose we restrict $\displaystyle \left| x - 1 \right| < 1 $ for example, then

$\displaystyle \begin{align*} -1 < x - 1 &< 1 \\
-3 < x - 3 &< -1 \end{align*}$

so we can say for certain that if $\displaystyle \left| x - 1 \right| < 1 $, then $\displaystyle \left| x - 3 \right| < 3 $. Therefore

$\displaystyle \begin{align*} \left| x - 1 \right| < \frac{2\,\epsilon}{7}\,\left| x - 3 \right| &< \frac{2\,\epsilon}{7} \cdot 3 \\
\left| x - 1 \right| < \frac{2\,\epsilon}{7}\,\left| x - 3 \right| &< \frac{6\,\epsilon}{7} \\
\left| x - 1 \right| &< \frac{6\,\epsilon}{7} \end{align*} $

So we can finally get to the proof now...

Let $\displaystyle \delta = \min \left\{ 1, \frac{6\,\epsilon}{7} \right\} $, then reverse every step and you are done.

Wish I knew how to do this stuff. Thanks Prove It.
 

FAQ: What is the Epsilon-Delta Method for Proving Limits?

What is an Epsilon-delta Limit Proof?

An Epsilon-delta Limit Proof is a method used to formally prove the limit of a function. It involves using two variables, epsilon and delta, to show that for every positive value of epsilon, there exists a positive value of delta that can be used to determine the behavior of the function near a specific point.

Why is an Epsilon-delta Limit Proof important?

An Epsilon-delta Limit Proof is important because it provides a rigorous and precise way to prove the limit of a function. This is especially useful in more advanced mathematics, where the concept of a limit is crucial in understanding the behavior of functions.

How does an Epsilon-delta Limit Proof work?

An Epsilon-delta Limit Proof works by using the definition of a limit, which states that for a given function f(x) and point c, the limit of f(x) as x approaches c is L if for every positive value of epsilon, there exists a positive value of delta such that if the distance between x and c is less than delta, then the distance between f(x) and L is less than epsilon.

What are some common challenges when using an Epsilon-delta Limit Proof?

Some common challenges when using an Epsilon-delta Limit Proof include finding the appropriate values of epsilon and delta, understanding the concept of a limit and how it relates to the function, and being able to manipulate the equations and inequalities involved in the proof.

Are there other methods for proving limits?

Yes, there are other methods for proving limits, such as using the Squeeze Theorem, the Intermediate Value Theorem, and L'Hôpital's rule. However, an Epsilon-delta Limit Proof is often considered the most rigorous and precise method for proving limits.

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