Limit of $\lim_{x\to 3} \frac{3-x}{6\sin \frac{\pi x}{18}-x}$

Question

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?

Answer 1

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 $\frac{\pi\sqrt3}{36}$.

Answer 2

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 $\dfrac{6}{6-\pi\sqrt{3}}$.

Answer 3

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.