Question

Determine whether f is continuous at c.

$f\left(x\right)=\left\{\begin{array}{l}\frac{x^2-2x}{x-2}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}x<2\\ 2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}x=2\\ \frac{x-4}{x-1}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}x>2\end{array}c=2\right.$

Short answer

f is not continuous at $c=2$.

Step-by-step solution

Step 1. Given information 

We have been given a function $f\left(x\right)=\left\{\begin{array}{l}\frac{x^2-2x}{x-2}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}x<2\\ 2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}x=2\\ \frac{x-4}{x-1}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}x>2\end{array}\right.$.

We have to determine whether this function is continuous at $c=2$.

Step 2. Write the condition for continuity

We say that f is continuous at c if it is defined at c, given that c is in the domain of f so that f(c) is equal to a number.

Also, another condition is when the right and left limits at c of the function f(x) are both equal to f(c).

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

However, in a case of a piecewise-defined function, different functions for different intervals must be taken into consideration.

Step 3. Find f(2).

Since$f\left(x\right)=\left\{\begin{array}{l}\frac{x^2-2x}{x-2}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}x<2\\ 2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}x=2\\ \frac{x-4}{x-1}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}x>2\end{array}\right.$

we get $f\left(2\right)=2$.

Step 4. Find left side limit. 

$$\begin{gathered} \underset{x\to 2^{-}}{lim} \left(\frac{x^2-2x}{x-2}\right) \\ =\underset{x\to 2^{-}}{lim} \left(\frac{x(x-2)}{x-2}\right) \\ =\underset{x\to 2^{-}}{lim} x \\ =2 \end{gathered}$$

Step 5. Find right-side limit. 

$$\begin{gathered} \underset{x\to 2^{+}}{lim} \left(\frac{x-4}{x-1}\right) \\ =\frac{\underset{x\to 2^{+}}{lim} x-\underset{x\to 2^{+}}{lim} 4}{\underset{x\to 2^{+}}{lim} x-\underset{x\to 2^{+}}{lim} 1} \\ =\frac{2-4}{2-1} \\ =\frac{-2}{1} \\ =-2 \end{gathered}$$

role="math" localid="1647085873721" $$\begin{gathered} \underset{x\to 2^{-}}{\lim }f\left(x\right)\ne \underset{x\to 2^{+}}{\lim }f\left(x\right) \\ \underset{x\to 2}{\lim }f\left(x\right)=f\left(2\right) \end{gathered}$$

Since, $\underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right)$

The function is not continuous at $c=2$.