Proving lim_{x->\frac{1}{10}}\frac{1}{x}=10

[Metal_Zelda]

Homework Statement

Prove lim_{x->\frac{1}{10}}\frac{1}{x}=10

Homework Equations

|f(x)-L|<epsilon, |x-a|<delta

The Attempt at a Solution

I need to go from 1/x to x, so I applied an initial condition of delta<1/20
<br /> \frac{-1}{20}&lt;x-\frac{1}{10}&lt;\frac{1}{20}
\frac{1}{10}&lt;x&lt;\frac{3}{10}
\frac{1}{x}&lt;10<br />

Moving on to the left side of the proof,
<br /> |\frac{1}{x}-10|&lt;\epsilon
-\epsilon&lt;\frac{1}{x}-10&lt;\epsilon

This is where I am stuck. If I use the fact that 1/x<10, I end up with -epsilon<10-10<epsilon, which isn't helpful.

 

[LCKurtz]

Homework Statement

Prove lim_{x-&gt;\frac{1}{10}}\frac{1}{x}=10

Homework Equations

|f(x)-L|<epsilon, |x-a|<delta

The Attempt at a Solution

I need to go from 1/x to x, so I applied an initial condition of delta<1/20
<br /> \frac{-1}{20}&lt;x-\frac{1}{10}&lt;\frac{1}{20}
\frac{1}{10}&lt;x&lt;\frac{3}{10}

I like that so far. Now look at your original problem$$
\left| \frac 1 x - 10\right| = \frac{|1-10x|}{|x|}= \frac{|\frac 1 {10}- x|}{|\frac x{10}|}$$which you are trying to make small. Can you underestimate the denominator using your last line above?

 

[Metal_Zelda]

Okay, so I took |1/x-10|<epsilon and did

\frac{1}{x}-\frac{10x}{x}&lt;\epsilon
\frac{1-10x}{x}&lt;\epsilon
\frac{-10(x-\frac{1}{10})}{x} &lt;\epsilon
-10(x-\frac{1}{10})&lt;\delta, \; \; \mbox{where} \; \; \delta=\frac{1}{20}

which leaves me with \frac{1}{200}&lt;x-\frac{1}{10}&lt;-\frac{1}{200}
,Which is impossible. epsilon/200 seems to satisfy the requirement, though.

Edit: Oops, I forgot to switch the inequality symbols. It should be

\frac{1}{200}&gt;x&gt;-\frac{1}{200}

 

[LCKurtz]

I wasn't done editing and accidentally posted. Look at it now.

 

[Metal_Zelda]

I like that so far. Now look at your original problem$$
\left| \frac 1 x - 10\right| = \frac{|1-10x|}{|x|}= \frac{|\frac 1 {10}- x|}{|\frac x{10}|}$$which you are trying to make small. Can you underestimate the denominator using your last line above?

I wasn't done editing and accidentally posted. Look at it now.

I made a few errors in my post as well. I'm going to retype everything I've done, please let me know if I've reached a solution.

|\frac{1}{x}-10|&lt;\epsilon \; \; \mbox{when} \; \; |x-\frac{1}{10}|&lt;\delta

Let δ=1/20. Then, expanding the left side, we get

-\frac{1}{20}&lt;x-\frac{1}{10}&lt;\frac{1}{20} \; \rightarrow \; \frac{1}{10}&lt;x&lt;\frac{3}{10} \; \iff \; \frac{1}{10}&lt;x \; \; \mbox {and} \; \; x&lt;\frac{3}{10}
Inverting the left side, we get
\frac{1}{x}&lt;10

Moving on to the right side,

<br /> \frac{1}{x}-\frac{10x}{x}&lt;\epsilon \; \rightarrow \; \frac{1-10x}{x}&lt;\epsilon \; \rightarrow \;<br /> \frac{-10(x-\frac{1}{10})}{x} &lt;\epsilon

We assume that
<br /> -10(x-\frac{1}{10})&lt;\delta, \; \; \mbox{where} \; \; \delta=\frac{1}{20}
<br /> \frac{1}{200}&gt;x-\frac{1}{10}&gt;-\frac{1}{200} \; \iff \; x-\frac{1}{10}&lt; \frac{1}{200}
Which implies that
\delta=min \{ \frac{1}{20},\frac{1}{200} \} \; \rightarrow \; \delta=\frac{1}{200}[\tex]<br /> <br /> I&#039;m assuming that I made a mistake on that last bit as 1/200 seems an odd bound.<br /> <br /> Edit: I&#039;m not sure why that last bit of latex isn&#039;t working. It should look like <a href="http://latex.codecogs.com/gif.latex?\delta=min&amp;space;\{&amp;space;\frac{1}{20},\frac{1}{200}&amp;space;\}&amp;space;\;&amp;space;\rightarrow&amp;space;\;&amp;space;\delta=\frac{1}{200}" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://latex.codecogs.com/gif.latex...ightarrow&amp;space;\;&amp;space;\delta=\frac{1}{200}</a>

 

[LCKurtz]

Your $\delta$ must depend on $\epsilon$. I think if you are careful you will find, with your original assumptions that $\delta =\frac \epsilon {200}$ will work. I suggest you work with absolute values. Think about underestimating $\frac {|x|}{10}$ in the denominator as I suggested. It's only another step or two.