<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] Calculate the limit of inverse:
$g(x)=\frac{6\sin\left(\frac{\pi x}{18}\right)-x}{3-x}=\frac{6\left(\sin\left(\frac{\pi x}{18}\right)-\frac12\right)+3-x}{3-x}$
$=\frac{6\left(\sin\left(\frac{\pi x}{18}\right)-\frac12\right)}{3-x}+1$
However
$\sin\left(\frac{\pi x}{18}\right)-\sin\left(\frac{\pi}{6}\right)=2\sin\left(\frac{\pi(x-3)}{36}\right)\cos\left(\frac{\pi(x+3)}{36}\right)$
So
$g(x)=\frac{12\sin\left(\frac{\pi(x-3)}{36}\right)\cos\left(\frac{\pi(x+3)}{36}\right)}{3-x}+1$
Using the limit $\lim_{u\to 0}\frac{\sin(u)}{u}=1$
$\lim_{x\to 3} \frac{\sin\left(\frac{\pi(x-3)}{36}\right)}{x-3}=\frac{\pi}{36}$
So $\lim_{x\to 3} g(x)=12.\left(-\frac{\pi}{36}\right)\cos\left(\frac{\pi}{6}\right)=1-\frac{\pi}{2\sqrt{3}}$
The result is the inverse.
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