1. Check $f(0)$ (Function Value)
The first part of the function is used for $x \geq 0$ :
$$
f(x)=\frac{\sqrt{9 x^4+x^2}}{5 x^2+3 x+1}
$$
Substitute $x=0$ :
$$
f(0)=\frac{\sqrt{9(0)^4+(0)^2}}{5(0)^2+3(0)+1}=\frac{\sqrt{0}}{1}=0
$$
$f(0)$ is defined and equal to $\mathbf{0}$ .
2. Check the Left-Hand Limit (LHL)
The second part of the function is used for $x<0$ :
$$
\begin{gathered}
L H L=\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} x \\
L H L=0
\end{gathered}
$$
3. Check the Right-Hand Limit (RHL)
The first part of the function is used for $x>0$ :
$$
R H L=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{9 x^4+x^2}}{5 x^2+3 x+1}
$$
Since this is a composite of a rational function and a square root, and the denominator is nonzero at $x=0$, we can substitute $x=0$ :
$$
R H L=\frac{\sqrt{9(0)^4+(0)^2}}{5(0)^2+3(0)+1}=\frac{\sqrt{0}}{1}=0
$$
- $f(0)=0$
- $\lim _{x \rightarrow 0} f(x)=0$
- $\lim _{z \rightarrow 0^{+}} f(x)=0$
Since $f(0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^*} f(x)=0$, all three conditions for continuity are met.
The function $f$ is continuous at $x=0$.
