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.


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

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

  • $\begingroup$ 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. $\endgroup$ – Zach Apr 12 '14 at 22:27
  • $\begingroup$ 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 $\lvert x-x_0\rvert\cdot b(x_0) < \epsilon$. $\endgroup$ – Daniel Fischer Apr 12 '14 at 22:32
  • $\begingroup$ 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 $\frac{1}{x^2{x_0}^2}<\frac{4}{{x_0}^4}$. $\endgroup$ – Zach Apr 13 '14 at 1:20
  • 1
    $\begingroup$ 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. $\endgroup$ – Zach Apr 13 '14 at 1:22
  • $\begingroup$ 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. $\endgroup$ – Daniel Fischer Apr 13 '14 at 1:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.