Evaluating $\lim_{x \to 1} \frac{\sin{\pi x}}{\sin{3\pi x}}$ without L'Hopital's rule

Question

I need help finding this limit:

$$\lim_{x \to 1} \frac{\sin{\pi x}}{\sin{3\pi x}}$$

I've thought of dividing and multiplying the numerator by $\pi x$ and the denominator by $3\pi x$ but it doesn't seem to get me anywhere.

Note that I'm not really familiar with L'Hopital's rule as of now, so any answers which do not include it would be much appreciated.

Answer 1

Recall that

$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$

We have

\begin{align*} \lim_{x \to 1} \frac{\sin{\pi x}}{\sin{3\pi x}}&=\lim_{x \to 1} \frac{\sin{\pi x}}{\sin(2\pi x+\pi x)}\\ &=\lim_{x \to 1} \frac{\sin{\pi x}}{\color{red}{\sin(2\pi x)}\cos(\pi x)+\cos(2\pi x)\sin(\pi x)}\\ &=\lim_{x \to 1} \frac{\sin{\pi x}}{\color{red}{2\sin(\pi x)\cos(\pi x)}\cos(\pi x)+\cos(2\pi x)\sin(\pi x)}\\ &=\lim_{x \to 1} \frac{1}{2\cos(\pi x)\cos(\pi x)+\cos(2\pi x)}\\ &=\frac{1}{2\cos(\pi)\cos(\pi)+\cos(2\pi)}\\ &=\frac13 \end{align*}

Answer 2

$\lim_{x \to 1} \frac{\sin{\pi x}}{\sin{3\pi x}}$

$=\lim_{x \to 1} \frac{\sin{\pi x}}{3\sin{\pi x}-4 \sin^3\pi x}$

$=\lim_{x \to 1} \frac{1}{\frac{3\sin{\pi x}-4 \sin^3{\pi x}}{\sin{\pi x}}}$

$=\lim_{x \to 1} \frac{1}{3-4 \sin^2(\pi x) }$

$=\frac{1}{3}$

Use, $\theta =\pi x $ in the formula $\sin3\theta=3\sin\theta- 4\sin^3{\theta}$

Answer 3

Since full answers have already been posted, I figured that it would be nice to show an alternative method which I believe was closer to your original attempt.

If the limit involved $x$ approaching $0$ rather than $1,$ then there are algebraic manipulations we could do to put the limit in terms of the special limit $\lim_{x \to 0} \frac{\sin x}{x} = 1.$ Thankfully, where defined this function is periodic with period $1$ (as the following manipulation will demonstrate) so we can achieve this with the substitution $u = x - 1 \leftrightarrow x = u + 1,$ and $u \to 0$:

$$\lim_{x \to 1} \frac{\sin (\pi x)}{\sin (3\pi x)} = \lim_{u \to 0} \frac{\sin(\pi u + \pi)}{\sin(3\pi u + 3\pi)} = \lim_{u \to 0} \frac{-\sin(\pi u)}{-\sin(3 \pi u)} = \lim_{u \to 0} \frac{\sin(\pi u)}{\sin(3 \pi u)}$$

taking advantage of the identity $\sin(\theta + \pi) = -\sin(\theta).$

Now we can proceed by rewriting the numerator and denominator in the form $\sin(\theta)/\theta.$

$$\lim_{u \to 0} \frac{\sin(\pi u)}{\sin(3 \pi u)} = \lim_{u \to 0} \frac{\sin(\pi u)}{\sin(3 \pi u)} \cdot \frac{3\pi u}{\pi u} \cdot \frac13 = \lim_{u \to 0} \frac{\sin(\pi u)/(\pi u)}{\sin(3 \pi u)/(3\pi u)} \cdot \frac13$$

Now, supposing for the moment that the limits all exist and that the denominator doesn't go to zero, we can split the limit over division as follows:

$$\lim_{u \to 0} \frac{\sin(\pi u)/(\pi u)}{\sin(3 \pi u)/(3\pi u)} \cdot \frac13 = \frac13 \cdot \frac{\lim_{u \to 0} \sin(\pi u)/(\pi u)}{\lim_{u \to 0} \sin(3\pi u)/(3\pi u)}$$

Now consider the substitutions $v_1 = \pi u$ and $v_2 = 3 \pi u$ for the numerator and denominator respectively. We now have

$$\frac13 \cdot \frac{\lim_{v_1 \to 0} \sin(v_1)/(v_1)}{\lim_{v_2 \to 0} \sin(v_2)/(v_2)} = \frac13 \cdot \frac11 = \boxed{\frac13}$$

As I indicated before, this approach is a bit more involved than the substitution, but I still find it pretty elegant and I thought you might like to see it.

Answer 4

Hint

$$f=\frac{\sin{(\pi x)}}{\sin{(3\pi x})}$$

$$x=y+1 \implies f=\frac{\sin{(\pi y)}}{\sin{(3\pi y})}=\frac 13\times\frac{\sin{(\pi y)}}{\pi y}\times\frac{3\pi y}{\sin{(3\pi y)}}$$ and $y \to 0$. Remember a very simple limit.

Answer 5

$$ \lim_{x \to {1}} \frac{\sin{\pi{x}}}{\sin{3\pi{x}}}. $$ To solve this problem I recommend to use the Taylor series: $$ \sin(x)=\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+....\right)=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!} .\tag{1} $$ $$ \sin({\alpha}+{\pi})=-\sin({\alpha}). \tag{2} $$ As Mr. Leibovici said, we need to perform a substitution $x=(y+1)$, therefore, $y\to{0}$.

Using the very well known formula $(2)$ and the substitution $x=(y+1)$ relatively our problem, therefore, we will have: $$ \lim_{y\to{0}} \frac{\sin{\pi{y}}}{\sin{3\pi{y}}}=\lim_{y\to{0}} \frac{\left(\pi{y}-\frac{(\pi{y})^3}{3!}+\frac{(\pi{y})^5}{5!}+\frac{(\pi{y})^7}{7!}+....\right)}{\left(3\pi{y}-\frac{(3\pi{y})^3}{3!}+\frac{(3\pi{y})^5}{5!}+\frac{(3\pi{y})^7}{7!}+....\right)}= \\ =\lim_{y\to{0}}\frac{y\left(\pi-\frac{\pi^{3}y^{2}}{3!}+\frac{\pi^{5}y^{4}}{5!}+\frac{\pi^{7}y^{6}}{7!}+....\right)}{y\left(3\pi-\frac{(3\pi)^{3}y^{2}}{3!}+\frac{(3\pi)^{5}y^{4}}{5!}+\frac{(3\pi)^{7}y^{6}}{7!}+....\right)}=\lim_{y\to{0}}\frac{\pi}{3\pi}=\frac{1}{3}. $$ Therefore, $$ \lim_{x \to {1}} \frac{\sin{\pi{x}}}{\sin{3\pi{x}}}=\frac{1}{3}. $$