Continuity

Decide whether or not the function

is continuous at $x=2$.

Answer:

If you try and evaluate the function $f\left(x\right)$ at 2, you get division by zero. So, in fact, this function is not defined at $x=2$, so it can't be continuous at that point either. You can graph the function to see that there is a vertical asymptote at $x=2$, which is why it isn't defined there.

This function is not continuous at x=2 | StudySmarter Originals

So the difficulty with the previous example was that the function wasn't defined when $x=2$. Suppose, instead, your function is defined by

$f\left(x\right)=\left\{\begin{array}{ll}\frac{x+2}{x-2},& x\ne 2\\ 7,& x=2\end{array}\right.$

which is definitely defined at $x=2.$ Is this function continuous at $x=2$?

Answer:

In this case

$\underset{x\to 2^{-}}{\lim }f\left(x\right)=-\infty$

but

$\underset{x\to 2+}{\lim }f\left(x\right)=\infty$,

so the limit doesn't exist at $x=2$. Therefore, even though the function is defined when $x=2$, it is not continuous there.

Let's rig the previous example some more. If the problem is that the limit from the left and right aren't the same, you can change the function a bit to see what happens. Take

$f\left(x\right)=\left\{\begin{array}{ll}\frac{x+2}{x-2},& x>2\\ 7,& x=2\\ -\frac{x+2}{x-2}& x<2\end{array}\right.$

Answer:

Now when you look at the limit as $x$ approaches 2, you have

$\underset{x\to 2^{-}}{\lim }f\left(x\right)=\infty =\underset{x\to 2+}{\lim }f\left(x\right).$

But $f\left(2\right)=7$, which is certainly not infinity! So the function is still not continuous at $x=2$.

Decide whether the function

is continuous when $x=3$.

Answer:

The trick here is to read the question carefully. We don't necessarily look at the point where the function changes definition, we look at what the question is asking!

In this case, the function changes definition at x=2, but we are asked if it is continuous at x=3. So all you need to consider is if the function $g\left(x\right)=x+3$ is continuous at $x=3$. But this is just a line, so you know

and the function $f\left(x\right)$ is continuous at $x=3$.

Let's work it out step by step:

Step 1: Make sure the function is defined at $x=3$. $g\left(3\right)=6$. Therefore, it is defined.

Step 2: Make sure the limit at $x=3$ exists. That is, check if the limit from the left and right of $x=3$ are equal.

$\underset{x\to 3^{-}}{\lim }g\left(x\right)=6$and $\underset{x\to 3^{+}}{\lim }g\left(x\right)=6$.

The limits from the left and right of $x=3$ are indeed equal.

Step 3: Lastly, check if the limit is equal to the function value at $x=3$. That is:

$\begin{array}{rcl}\underset{x\to 3}{\lim }g\left(x\right)& =& g\left(3\right)\\ & & \\ 6& =& 6\end{array}$

This last condition is satisfied. Therefore, the function is continuous at $x=3$.

Decide whether the function

$f\left(x\right)=\left\{\begin{array}{ll}-x^2+9,& x\le 2\\ x+3,& x>2\end{array}\right.$

is continuous at the point $p=2$.

Answer:

Let's follow the same steps as the previous example.

Step 1: Check if the function is defined at $p=2$.

$f\left(2\right)=-2^2+9=-4+9=5$.

Therefore, the function is defined at $p=2$.

Step 2: Now you check if the limit exists. The limit from the left gives

$\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{-}}{\lim }(-x^2+9)=5,$

and the limit from the right is

$\underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }(x+3)=5,$

which means that

$\underset{x\to 2}{\lim }f\left(x\right)=5.$

Step 3: Lastly, check if the function value from step 1 and limit from step 2 are equal. They both equal 5!

So, you've checked:

1. That the function is defined at the point,
2. The limit of the function exists at that point, and
3. The function value at that point has the same value as the limit.

Therefore, the function $f\left(x\right)$ is continuous at $p=2$.

Decide whether the function

$f\left(x\right)=\left\{\begin{array}{ll}-x^2+9,& x\le 2\\ x+1,& x>2\end{array}\right.$

is continuous at the point $p=2$.

Answer:

Step 1:

Just like in the previous example, $f\left(2\right)=5$

Step 2: Check left and right-hand limits:

The limit from the left:

$\underset{x\to 2^{-}}{\lim }f\left(x\right)=5.$

But now the limit from the right is

$\underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }(x+1)=2,$

so

$\underset{x\to 2}{\lim }f\left(x\right)$

does not exist. Therefore the function is not continuous at the point $p=2$.