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I'm trying to prove that $\lim_{x \to x_0} \frac{1}{ x^2 } = \frac{1}{ {x_0}^2 }$. I know this means that for all $\epsilon > 0$, I must show that there exists a $\delta > 0$ such that $\left | x - x_0 \right | < \delta \Rightarrow \left | \frac{1}{ x^2 } - \frac{1}{ {x_0}^2 } \right | < \epsilon$. Since $f(x) = \frac{1}{x^2}$ is an even function, I'll restrict the domain to $x>0$. After some manipulation of $\left | \frac{1}{ x^2 } - \frac{1}{ {x_0}^2 } \right | < \epsilon$ I have $\left | x - x_0 \right | < \epsilon\frac{x^2{x_0}^2}{x+{x_0}}$. I know that I can't just let $\delta = \epsilon\frac{x^2{x_0}^2}{x+{x_0}}$ since $\delta$ should not depend on $x$, so I need something else. This is where I'm stuck. I think I need to choose an upper bound for $\left|x-x_0\right|$ which depends on $x_0$ and find a sufficient $\delta$ from this restriction, but I'm not sure how to go about this.

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Take several steps.

First, choose a $\delta_0 > 0$ such that for $\lvert x-x_0\rvert \leqslant \delta_0$ you can bound the expression $\dfrac{x+x_0}{x^2x_0^2}$ by an expression depending only on $x_0$. A choice of the form $\delta_0 = c\cdot x_0$ is the canonical way.

Then use the bound involving only $x_0$ - let's call it $b(x_0)$ - to get a $\delta_1 > 0$, namely $\delta_1 = \dfrac{\epsilon}{b(x_0)}$, and choose $\delta := \min \{ \delta_0,\delta_1\}$ to obtain the desired

$$\lvert x-x_0\rvert < \delta \implies \left\lvert \frac{1}{x^2} - \frac{1}{x_0^2}\right\rvert < \epsilon.$$

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Do you instead mean "...bound the expression $\frac{x^2x_0^2}{x+x_0}$ by an expression..."? I'm not sure why you took the reciprocal. – Zach Apr 12 '14 at 22:27
Because you have $$\left\lvert \frac{1}{x^2} - \frac{1}{x_0^2}\right\rvert = \lvert x-x_0\rvert \cdot \frac{x+x_0}{x^2x_0^2},$$ and that means you want an upper bound $b(x_0)$ on $\frac{x+x_0}{x^2x_0^2}$, so that $\lvert x-x_0\rvert\cdot b(x_0) < \epsilon$. – Daniel Fischer Apr 12 '14 at 22:32
Let me check my answer in two parts. Suppose $\lim_{x \to x_0} \frac{1}{x^2} = \frac{1}{{x_0}^2}$. Then $\forall\epsilon>0$ $\exists\delta>0$ such that $\forall x>0$ $\left|x-x_0\right|<\delta \Rightarrow \left|\frac{1}{x^2}-\frac{1}{{x_0}^2}\right|<\epsilon$. Let $\delta_1 = \frac{1}{2}x_0$. Then $\left|x-x_0\right|<\delta_1=\frac{1}{2}x_0$, hence $\frac{1}{2}x_0<x<\frac{3}{2}x_0$. Then $\frac{1}{2}x_0<x$, hence $\frac{1}{2}{x_0}^2<xx_0$, $\frac{1}{4}{x_0}^4<x^2{x_0}^2$, and $\frac{1}{x^2{x_0}^2}<\frac{4}{{x_0}^4}$. – Zach Apr 13 '14 at 1:20
Also $x<\frac{3}{2}x_0$, hence $x+x_0<\frac{3}{2}x_0+x_0=\frac{5}{2}x_0$. Then $\frac{x+x_0}{x^2{x_0}^2}<\frac{4}{{x_0}^4}\frac{5x_0}{2}=\frac{10}{{x_0}^3}$, and $\left|\frac{1}{x^2}-\frac{1}{{x_0}^2}\right|=\left|x-x_0\right|\frac{x+x_0}{x^2‌​{x_0}^2}<\left|x-x_0\right|\frac{10}{{x_0}^3}<\epsilon$. Let $\delta_2=\epsilon\frac{{x_0}^3}{10}$ and let $\delta=\min(\delta_1,\delta_2)$. Thus $\left|x-x_0\right|<\delta \Rightarrow \left|\frac{1}{x^2}-\frac{1}{{x_0}^2}\right|<\epsilon$. Therefore the supposition is true. – Zach Apr 13 '14 at 1:22
Right. Only you wrote $\lvert x-x_0\rvert \frac{10}{x_0^3} < \epsilon$ too early, you only know that after having introduced $\delta_2$ and constrained $\lvert x-x_0\rvert < \min\{\delta_1,\delta_2\}$. Perhaps you meant $\lvert x-x_0\rvert \frac{10}{x_0^3} \overset{!}{<} \epsilon$ as a requirement and not the inequality as a proposition, though. – Daniel Fischer Apr 13 '14 at 1:28

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