Determine whether f is continuous at c.

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

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

We have to determine whether this function is continuous at $c=1$.

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.

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

we get $f\left(1\right)=2$.

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

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

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

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

The function is not continuous at $c=1$.