Determine whether f is continuous at c.

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

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

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

we get $f\left(0\right)=2$.

$$\begin{gathered} \underset{x\to 0^{-}}{lim} 2e^x \\ =2\underset{x\to 0^{-}}{lim} e^x \\ =2e^0 \\ =2 \end{gathered}$$

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

role="math" localid="1647086351667" $$\begin{gathered} \underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right) \\ \underset{x\to 0}{\lim }f\left(x\right)=2 \end{gathered}$$

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

The function is continuous at $c=0$.