Question

Determine whether f is continuous at c.

$f\left(x\right)=\left\{\begin{array}{l}\frac{x^3+3x}{x^2-3x}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}x\ne 0\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}x=0\end{array}c=0\right.$

Short answer

f is not continuous at $c=0$.

Step-by-step solution

Step 1. Given information 

We have been given a function $f\left(x\right)=\left\{\begin{array}{l}\frac{x^3+3x}{x^2-3x}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}x\ne 0\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}x=0\end{array}\right.$.

We have to determine whether this function is continuous at $c=0$.

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(0)

Since $f\left(x\right)=\left\{\begin{array}{l}\frac{x^3+3x}{x^2-3x}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}x\ne 0\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}x=0\end{array}\right.$

we get$f\left(0\right)=1$

Step 4. Find left side limit.

$$\begin{gathered} \underset{x\to 0^{-}}{lim} \left(\frac{x^3+3x}{x^2-3x}\right) \\ =\underset{x\to 0^{-}}{lim} \left(\frac{x\left(x^2+3\right)}{x(x-3)}\right) \\ =\underset{x\to 0^{-}}{lim} \left(\frac{x^2+3}{x-3}\right) \\ =\frac{\underset{x\to 0^{-}}{lim} x^2+\underset{x\to 0^{-}}{lim} 3}{\underset{x\to 0^{-}}{lim} x-\underset{x\to 0^{-}}{lim} } \\ =\frac{0^2+3}{0-3} \\ =\frac{3}{-3} \\ =-1 \end{gathered}$$

Step 5. Find right-side limit.

_{$$\begin{gathered} \underset{x\to 0^{+}}{lim} \left(\frac{x^3+3x}{x^2-3x}\right) \\ =\underset{x\to 0^{+}}{lim} \left(\frac{x\left(x^2+3\right)}{x(x-3)}\right) \\ =\underset{x\to 0^{+}}{lim} \left(\frac{x^2+3}{x-3}\right) \\ =\frac{\underset{x\to 0^{+}}{lim} x^2+\underset{x\to 0^{+}}{lim} 3}{\underset{x\to 0^{+}}{lim} x-\underset{x\to 0^{+}}{lim} } \\ =\frac{0^2+3}{0-3} \\ =\frac{3}{-3} \\ =-1 \end{gathered}$$}

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)$

$\underset{x\to 0}{lim}f\left(x\right)=-1$

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

The function is not continuous at $c=0$.