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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?

Thanks,

~~Bart

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    $\begingroup$ 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
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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 $x = 0$.

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  • $\begingroup$ 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

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