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?

I assume you meant elementary limits are allowed. I'll use:
$$\lim_{x\to 0} \frac{\sin x}{x}=1$$
Instead of the fraction, it's easier to look at its inverse:
$$
\begin{aligned}
\lim_{x\to 3} \frac{6\sin \frac{\pi x}{18}-x}{3-x}&=1+\lim_{x\to 3}\frac{6\sin \frac{\pi x}{18}-3}{3-x}\\
&= 1+ 6\lim_{x\to 3}\frac{\sin \frac{\pi x}{18}-\sin \frac{\pi}{6}}{3-x}\\
&= 1+ 12\lim_{x\to 3}\frac{\sin \left(\frac{\pi x}{36}-\frac{\pi}{12}\right)\cos \left(\frac{\pi x}{36}+\frac{\pi}{12}\right)}{3-x}\\
&= 1+6\sqrt{3}\lim_{x\to 3}\frac{\sin \left(\frac{x-3}{36}\pi\right)}{3-x}\\
&= 1-6\sqrt{3}\lim_{x\to 3}\frac{\sin \left(\frac{\pi}{36}(x-3)\right)}{\frac{\pi}{36}(x-3)} \cdot \frac{\pi}{36} \\
&=1-\frac{\pi\sqrt{3}}{6}
\end{aligned}
$$
Therefore your limit equals $\dfrac{6}{6-\pi\sqrt{3}}$.
