Yesenia Obrien

2022-07-03

I need help finding this limit:
$\underset{x\to 1}{lim}\frac{\mathrm{sin}\pi x}{\mathrm{sin}3\pi x}$

Caiden Barrett

Expert

Recall that
$\mathrm{sin}\left(A+B\right)=\mathrm{sin}A\mathrm{cos}B+\mathrm{cos}A\mathrm{sin}B$
We have
$\begin{array}{rl}\underset{x\to 1}{lim}\frac{\mathrm{sin}\pi x}{\mathrm{sin}3\pi x}& =\underset{x\to 1}{lim}\frac{\mathrm{sin}\pi x}{\mathrm{sin}\left(2\pi x+\pi x\right)}\\ & =\underset{x\to 1}{lim}\frac{\mathrm{sin}\pi x}{\mathrm{sin}\left(2\pi x\right)\mathrm{cos}\left(\pi x\right)+\mathrm{cos}\left(2\pi x\right)\mathrm{sin}\left(\pi x\right)}\\ & =\underset{x\to 1}{lim}\frac{\mathrm{sin}\pi x}{2\mathrm{sin}\left(\pi x\right)\mathrm{cos}\left(\pi x\right)\mathrm{cos}\left(\pi x\right)+\mathrm{cos}\left(2\pi x\right)\mathrm{sin}\left(\pi x\right)}\\ & =\underset{x\to 1}{lim}\frac{1}{2\mathrm{cos}\left(\pi x\right)\mathrm{cos}\left(\pi x\right)+\mathrm{cos}\left(2\pi x\right)}\\ & =\frac{1}{2\mathrm{cos}\left(\pi \right)\mathrm{cos}\left(\pi \right)+\mathrm{cos}\left(2\pi \right)}\\ & =\frac{1}{3}\end{array}$

aggierabz2006zw

Expert

$\underset{x\to 1}{lim}\frac{\mathrm{sin}\pi x}{\mathrm{sin}3\pi x}.$
To solve this problem I recommend to use the Taylor series:
$\begin{array}{}\text{(1)}& \mathrm{sin}\left(x\right)=\left(x-\frac{{x}^{3}}{3!}+\frac{{x}^{5}}{5!}+\frac{{x}^{7}}{7!}+....\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}{x}^{2n+1}}{\left(2n+1\right)!}.\end{array}$
$\begin{array}{}\text{(2)}& \mathrm{sin}\left(\alpha +\pi \right)=-\mathrm{sin}\left(\alpha \right).\end{array}$
we need to perform a substitution $x=\left(y+1\right)$, therefore, $y\to 0$
Using the very well known formula (2) and the substitution $x=\left(y+1\right)$ relatively our problem, therefore, we will have:
$\underset{y\to 0}{lim}\frac{\mathrm{sin}\pi y}{\mathrm{sin}3\pi y}=\underset{y\to 0}{lim}\frac{\left(\pi y-\frac{\left(\pi y{\right)}^{3}}{3!}+\frac{\left(\pi y{\right)}^{5}}{5!}+\frac{\left(\pi y{\right)}^{7}}{7!}+....\right)}{\left(3\pi y-\frac{\left(3\pi y{\right)}^{3}}{3!}+\frac{\left(3\pi y{\right)}^{5}}{5!}+\frac{\left(3\pi y{\right)}^{7}}{7!}+....\right)}=\phantom{\rule{0ex}{0ex}}=\underset{y\to 0}{lim}\frac{y\left(\pi -\frac{{\pi }^{3}{y}^{2}}{3!}+\frac{{\pi }^{5}{y}^{4}}{5!}+\frac{{\pi }^{7}{y}^{6}}{7!}+....\right)}{y\left(3\pi -\frac{\left(3\pi {\right)}^{3}{y}^{2}}{3!}+\frac{\left(3\pi {\right)}^{5}{y}^{4}}{5!}+\frac{\left(3\pi {\right)}^{7}{y}^{6}}{7!}+....\right)}=\underset{y\to 0}{lim}\frac{\pi }{3\pi }=\frac{1}{3}.$
Therefore,
$\underset{x\to 1}{lim}\frac{\mathrm{sin}\pi x}{\mathrm{sin}3\pi x}=\frac{1}{3}.$

Do you have a similar question?