Determine whether f is continuous at c.

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

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

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

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$f\left(x\right)=\left\{\begin{array}{l}3\cos x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}x<0\\ 3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}x=0\\ \frac{x^3+3x^2}{x^2}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}x>0\end{array}\right.$

we get role="math" localid="1647086638967" $f\left(0\right)=3$.

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

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

role="math" localid="1647086880582" $$\begin{gathered} \underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right) \\ \underset{x\to 0}{\lim }f\left(x\right)=3 \end{gathered}$$

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

The function is continuous at $c=0$.