# Prove the Limit

1. Mar 19, 2009

### H2Pendragon

1. The problem statement, all variables and given/known data
Use the definition of limits to show that
$$lim _{x\rightarrow 2} \frac{1}{x^{2}} = \frac{1}{4}$$

2. Relevant equations
$$\forall \epsilon > 0, \exists \delta > 0, x \in D, 0 < |x-2| < \delta \Rightarrow |\frac{1}{x^{2}} - \frac{1}{4}| < \epsilon$$

3. The attempt at a solution

$$|\frac{1}{x^{2}} - \frac{1}{4}| < \epsilon \Rightarrow |\frac{4-x^{2}}{4x^{2}}| < \epsilon \Rightarrow \frac{|4-x^{2}|}{4x^{2}} < \epsilon$$

$$Restrict: 1 < x < 3 \Rightarrow 4<4x^{2}<36 \Rightarrow \frac{1}{36} < \frac{1}{4x^{2}} < \frac{1}{4} \Rightarrow \frac{|4-x^{2}|}{36} < \frac{|4-x^{2}|}{4x^{2}} < \frac{|4-x^{2}|}{4} < \epsilon$$

And now I'm stuck. I'm trying to find delta based on epsilon. Normally the problems come to a neat little solution, but |4-x2| is not |x-2| and trying to get it to equal |x-2| is where I'm having the trouble.

Any help is appreciated.

2. Mar 19, 2009

### gabbagabbahey

$|4-x^2|=|x^2-4|=|(x-2)(x+2)|=$____?

3. Mar 19, 2009

### H2Pendragon

Yeah I got that far. I know that
$$|(x-2)||(x+2)| < 4\epsilon$$

But that still leaves that (x+2) there (which I can remove the absolute value from since it has to be positive). I can't have delta dependent on x.

I'm probably missing something blatant and obvious right here.

Last edited: Mar 19, 2009
4. Mar 19, 2009

### gabbagabbahey

Well, you've restricted your domain to between 1 and 3, so..... __?__$\leq|x+2|\leq$__?__ .....then simply choose epsilon to be greater than the upper bound of $$\frac{|x-2||x+2|}{4}$$

5. Mar 19, 2009

### H2Pendragon

ah, duh, thanks. 3 < |x+2| < 5 (since x+2 is positive with or without the absolute value)

So then

$$\frac{|x-2||x+2|}{4} \leq \frac{5*|x-2|}{4} < \epsilon$$

Thus $$|x-2| < \frac{4}{5}\epsilon$$

Thus if $$\delta = \frac{4}{5}\epsilon \Rightarrow |f(x) - \frac{1}{4}| < \epsilon$$

Thanks so much.

6. Mar 20, 2009

1<x<3
implies
3<x+2<5