MHB What is the Epsilon-Delta Method for Proving Limits?

[nycmathdad]

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...

 

[Prove It]

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.

 

[jonah1]

Beer soaked ramblings follow.

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.

 

[nycmathdad]

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.