How do I tell if a function is continuous?

Question

Is the function $f(x)=9-x^2$ continuous?

1.Lets say for $x=1$

$f(1)= 9-1=8$

2.$\lim\limits_{x \to 1} (9-x^2)= 9-1 = 8$

3.So $f(a) = \lim\limits_{x \to a} f(x) $

Does this mean that the function is continuous ?

Answer 1

Yes, you've got the right steps for the continuity of $f$ at the point $x = 1$. To generalise it to any point $x = a$, change $1$ to $a$ in the above steps.

1. $f(a) = 9-a^2$
2. $\lim\limits_{x \to a} f(x) = \lim\limits_{x \to a} (9 - x^2) \stackrel{(*)}{=} 9-a^2$
3. so $f(a) = \lim\limits_{x \to a} f(x)$.

Since the choice of $a \in \Bbb R$ is arbitrary, $f$ is continuous on $\Bbb R$.

We have (*) since for any $\epsilon > 0$, we take $\delta = \min((2|a|+1)^{-1} \epsilon, 1)$. For any $x \in (a-\delta,a+\delta)$, $|x-a| < \delta$ and $|x+a| \le |x| + |a| < (|a| + \delta) + |a| \le 2|a| + 1$. $$|(9-x^2)-(9-a^2)| = \dots = |x+a||x-a| < (2|a|+1)\delta < \epsilon$$

Answer 2

I don't want to explicitly walk you through the answer because this looks like a homework question and it's important that the solution you come up with is your own so I'm only going to provide you with some direction.

In regards to the steps you walked through, you proved that a limit exists at $x=1$, which as a result means that $f(x)$ continuous at $x=1$, which is true. However, your steps only prove that $f(x)$ is continuous at $x=1$ and not for any domain $D$.

For that, you will need to prove that for every $n \in D$ a limit exists at $n$, or you could show that the derivative of $f(x)$ is defined in the domain $D$.

To prove that for every $n \in D$ a limit exists at $n$, you just need to prove the following,

$$\lim\limits_{x \to n} f(x) = f(n)$$

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As a side note, this informal definition of continuity may help you grasp the concept a little better.

If you can draw the graph of $f(x)$ along the domain $D$ without taking your pencil off the paper, then $f(x)$ is continuous along $D$.