I am helping my sister with Calculus i and I am stumped by this problem:

Is the following function discontinuous?

$f(x)=\left\{\begin{aligned} &\frac{1}{x^2} &&if\ x\ne0 \\ &1 &&if\ x=2\end{aligned} \right.$

The textbook gives the short explanation:

Here $\ f(0)=1$ is defined but

$\lim_{x\to 0} f(x)=\lim_{x\to 0} \frac{1}{x^2}$

does not exist. So $f$ is discontinuous at zero.

I don't understand. What does not exist? The statement

$\lim_{x\to 0} f(x)=\lim_{x\to 0} \frac{1}{x^2}$

or just one of the two limits? By my thinking,

$\lim_{x\to 0} \frac{1}{x^2}=1$ and $\lim_{x\to 0} 1=1$

so therefore I would think that

$\lim_{x\to 0} f(x)=1$.

Would I be wrong in thinking that?



  • 1
    $\begingroup$ As $x$ gets closer to $0$, $\frac{1}{x^2}$ becomes really large. So, the limit does not exist. $\endgroup$ – Joe Johnson 126 Aug 30 '17 at 23:58

If we look at the behaviour as $x$ approaches zero from the right, the function looks like this:

$$\begin{matrix}x & f(x) = \frac{1}{x^2} \\ 1 & 1 \\ 0.1 & 100 \\ 0.01 & 10000 \\ 0.001 & 1000000 \\ 0.0001 & 100000000\end{matrix}$$

Notice how as $x$ gets smaller and smaller, $f(x)$ gets bigger and bigger. If we do the same thing from the left (i.e. trying very small negative $x$ values), then the same thing happens - for example, $f(-0.001) = 1000000$. So $\lim_{x \rightarrow 0^+}f(x) = \infty = \lim_{x \rightarrow 0^-}f(x)$, and an infinite limit means the function cannot be continuous at $x = 0$.

  • $\begingroup$ Upvote, but since it is $x^2$ you don't get negative infinity on the left side. $\endgroup$ – Joe Johnson 126 Aug 31 '17 at 0:06
  • $\begingroup$ Dang. I cannot brain today. Let me just surreptitiously fix that. $\endgroup$ – ConMan Aug 31 '17 at 4:39

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.