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**1. The problem statement, all variables and given/known data**

Use the definition of limits to show that

[tex]lim _{x\rightarrow 2} \frac{1}{x^{2}} = \frac{1}{4}[/tex]

**2. Relevant equations**

[tex]\forall \epsilon > 0, \exists \delta > 0, x \in D, 0 < |x-2| < \delta \Rightarrow |\frac{1}{x^{2}} - \frac{1}{4}| < \epsilon [/tex]

**3. The attempt at a solution**

[tex]|\frac{1}{x^{2}} - \frac{1}{4}| < \epsilon \Rightarrow |\frac{4-x^{2}}{4x^{2}}| < \epsilon \Rightarrow \frac{|4-x^{2}|}{4x^{2}} < \epsilon [/tex]

[tex]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[/tex]

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-x

^{2}| is not |x-2| and trying to get it to equal |x-2| is where I'm having the trouble.

Any help is appreciated.