Answer
The Maclaurin series for $f(x)=\sin(x^2)$ is convergent for any value of $x$.
Work Step by Step
By making use of Table 2, we have the Maclaurin series for $f(x)=\sin x$ as follows
$$f(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$$
Now by the comparison with the function $f(x)=\sin x^2$, we have the Maclaurin series as follows
$$f(x)=\sin x^2= x^2-\frac{(x^2)^3}{3!}+\frac{(x^2)^5}{5!}-\frac{(x^2)^7}{7!}+...\\
=x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\frac{x^{14}}{7!}+...$$
Moreover, again from Table 2, the Maclaurin series for $f(x)=\sin x$ is convergent for any value of $x$. Hence the Maclaurin series for $f(x)=\sin(x^2)$ is also convergent for any value of $x$.
