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I would like to prove that $$ \lim_{x \rightarrow 2}\frac{1}{x^2}=\frac{1}{4} $$ using the epsilon-delta definition of a limit.

I start with the formal definition:

For every $\varepsilon$ > 0 there exist a $\delta$ > 0 such that if $|x - a| < \delta$, then $|f(x) - a|$ < $\varepsilon$.

So I need to prove that if $|x - 2| < \delta$ then $|\frac{1}{x^2} - \frac{1}{4}| < \varepsilon$ .

I start by doing scratch-work to turn the left hand side of $|\frac{1}{x^2} - \frac{1}{4}| < \varepsilon$ into $|x-2|$ $$ |\frac{1}{x^2} - \frac{1}{4}| < \varepsilon $$ $$ |\frac{4-x^2}{4x^2}| < \varepsilon $$ $$ |4-x^2| < 4x^2\varepsilon $$ $$ |(x+2)(x-2)| < 4x^2\varepsilon $$ $$ |x-2| < \frac{4x^2}{|x+2|}\varepsilon $$ I would be able to complete the proof if the right-hand side were simply $\frac{\varepsilon}{|x+2|}$, but the $4x^2$ is throwing me off.

Normally I would pick: $\delta \leq 1$ then use $|x-2|<\delta=1$, or $-1<x-2<1$ to turn $x-2$ into the denominator with $\varepsilon$, then pick the lowest bound of the inequality, giving me $\delta = min:\{1, \frac{\varepsilon}{<lowest bound>}\}$

I don't know how to handle the extra $4x^2$ however. Any help would be appreciated.

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Consider instead backing up a bit to the point where you had the expression $$\left\vert \frac{4 - x^2}{4x^2} \right\vert.$$ If you choose $\delta < 1$, then $\vert 4x^2 \vert > 4$, hence we have $$\left\vert \frac{4 - x^2}{4x^2} \right\vert < \frac{\vert (2 - x)(2 + x) \vert}{4}.$$ Can you finish now?

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