<s>[INST] Limit of $\lim_{x\to 3} \frac{3-x}{6\sin \frac{\pi x}{18}-x}$ I want to solve a limit without l'Hospital, just with algebraic manipulation:
$$\lim_{x\to 3} \frac{3-\sqrt{6+x}}{6\sin \frac{\pi x}{18}-x}$$
I started with:
$$\lim_{x\to 3} \frac{3-\sqrt{6+x}}{6\sin \frac{\pi x}{18}-x}=\lim_{x\to 3} \frac{(3-\sqrt{6+x})(3+\sqrt{x+6})}{(6\sin \frac{\pi x}{18}-x)(3+\sqrt{x+6})}=\lim_{x\to 3} \frac{3-x}{(6\sin \frac{\pi x}{18}-x)(3+\sqrt{x+6})}$$
$3+\sqrt{x+6} \to 6$ is determinate, so I only need to calculate:
$$\lim_{x\to 3} \frac{3-x}{6\sin \frac{\pi x}{18}-x}$$
but I could only do it with l'Hospital:
$$\lim_{x\to 3} \frac{3-x}{6\sin \frac{\pi x}{18}-x}=\lim_{x\to 3} \frac{-1}{6\cdot \frac{\pi}{18}\cos \frac{\pi x}{18}-1}=\frac{-1}{\frac{\pi\sqrt{3}}{6}-1}$$
Can I get some help without l'Hospital?
 [/INST] Clearly, it is enough to compute$$\lim_{x\to3}\frac{\sin\left(\frac{\pi x}{18}\right)-\frac x6}{x-3}\tag1$$and, in order to compute that, it will be enough to compute$$\lim_{x\to3}\frac{\sin\left(\frac{\pi x}{18}\right)-\frac 12}{x-3}\text{ and }\lim_{x\to3}\frac{\frac x6-\frac12}{x-3},\tag2$$since $(1)$ is equal to$$\lim_{x\to3}\frac{\sin\left(\frac{\pi x}{18}\right)-\frac 12}{x-3}-\lim_{x\to3}\frac{\frac x6-\frac12}{x-3}$$(assuming that both limits from $(2)$ exist). But, clearly,$$\lim_{x\to3}\frac{\frac x6-\frac12}{x-3}=\lim_{x\to3}\frac16\frac{x-3}{x-3}=\frac16,$$whereas$$\lim_{x\to3}\frac{\sin\left(\frac{\pi x}{18}\right)-\frac 12}{x-3}$$is the derivative at $3$ of $x\mapsto\sin\left(\frac{\pi x}{18}\right)$, which is equal to $\frac{\pi\sqrt3}{36}$.
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