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
$$\frac{-1}{20}<x-\frac{1}{10}<\frac{1}{20}$$
$$\frac{1}{10}<x<\frac{3}{10}$$
$$\frac{1}{x}<10$$

Moving on to the left side of the proof,
$$|\frac{1}{x}-10|<\epsilon$$
$$-\epsilon<\frac{1}{x}-10<\epsilon$$

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

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
$$\frac{-1}{20}<x-\frac{1}{10}<\frac{1}{20}$$
$$\frac{1}{10}<x<\frac{3}{10}$$

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

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

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

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

$$\frac{1}{200}>x>-\frac{1}{200}$$

 

[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.

$$|\frac{1}{x}-10|<\epsilon \; \; \mbox{when} \; \; |x-\frac{1}{10}|<\delta$$

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

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

Moving on to the right side,

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

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

 

[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.