## University Calculus: Early Transcendentals (3rd Edition)

By $c=9$ will $f(x)$ be continuous at $x=0$.
$f(x)=\frac{\sin^23x}{x^2}$ for $x\ne0$ and $f(x)=c$ for $x=0$. $f(x)$ is continuous at $0$ if and only if $\lim_{x\to0}f(x)=f(0)=c$ First, we need to find $\lim_{x\to0}f(x)$: $$\lim_{x\to0}f(x)=\lim_{x\to0}\frac{\sin^23x}{x^2}=\lim_{x\to0}\Big(\frac{\sin^23x}{3x^2}\times3\Big)$$ $$\lim_{x\to0}f(x)=\Big(3\lim_{x\to0}\frac{\sin3x}{3x}\Big)^2$$ Recall that $\lim_{a\to0}\frac{\sin a}{a}=1$. With $a=3x$ here, we still can apply the identity: $$\lim_{x\to0}f(x)=(3\times1)^2=9$$ As stated above, for $f(x)$ to be continuous at $0$, $f(0)=c$ needs to acquire the same value as $\lim_{x\to0}f(x)$. And because $\lim_{x\to0}f(x)=9$, by $c=9$ will $f(x)$ be continuous at $x=0$.