Finding limit of $\frac{\sin\pi3^x}{x}$ as $x\to 0$ I have to find the limit of $\frac{\sin\pi3^x}{x}$ as $x\to 0$ using ONLY notable limits
please help me.
 A: We can use the fact that $\sin\alpha=-\sin(\alpha-\pi)$, so we can rewrite the limit as
$$
\lim_{x\to0}-\frac{\sin(\pi(3^x-1))}{x}=
-\lim_{x\to0}\frac{\sin(\pi(3^x-1))}{\pi(3^x-1)}\frac{\pi(3^x-1)}{x}
$$
The limits of the two fractions are known limits. The original limit is $-\pi\log3$.
Using the chain rule for computing the derivative of the numerator is much more instructive, in my opinion.
A: For $x\to0, \; \operatorname{e}^x\sim 1+x+o(x^2)$ so that $3^x\sim 1+x\ln 3$
$$
\frac{\sin(\pi \operatorname{3}^x)}{x}\sim
\frac{\sin(\pi +\pi x\ln 3)}{x}=-\pi\ln3\frac{\sin(\pi x\ln 3)}{\pi x\ln 3}\to-\pi\ln3
$$
A: $\sin x=\sin(\pi-x)\quad,\quad\lim_{x\to0}\frac{\sin x}x=1\quad,\quad\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f'(a)\iff$ $$\lim_{x\to0}\frac{\sin(\pi\cdot3^x)}x=\lim_{x\to0}\frac{\sin(\pi-\pi\cdot3^x)}x=\lim_{x\to0}\frac{\sin(\pi-\pi\cdot3^x)}{\pi-\pi\cdot3^x}\cdot\frac{\pi-\pi\cdot3^x}{x}=$$ $$=\pi\cdot\lim_{x\to0}1\cdot\frac{1-3^x}x=-\pi\cdot\lim_{x\to0}\frac{3^x-3^0}{x-0}=-\pi\cdot(3^x)'_{x=0}=-\pi\cdot(3^x\cdot\ln3)_{x=0}=-\pi\cdot\ln3.$$
