Continuity practice question 5

Which piece gives $f(0)$, and which piece do you need a limit of? What happens if you put $x = 0$ straight into that one?

Continuity at $0$ needs $\lim_{x \to 0^{-}} f(x) = f(0) = c$. The left piece gives $\tfrac{0}{0}$ at $x = 0$, so rewrite it using $\lim_{u \to 0} \dfrac{\sin u}{u} = 1$.

Multiply and divide by $6x$, so the denominator matches the argument of sine: $\dfrac{\sin(6x)}{2x} = \dfrac{6}{2}\cdot\dfrac{\sin(6x)}{6x}$.