michiiiiiakqm

2022-03-28

proving that $\underset{x\to \pi }{lim}\frac{1+\mathrm{cos}x}{1+\mathrm{cos}3x}=\frac{1}{9}$ without L'Hôpital

Avery Maxwell

As $\mathrm{cos}\left(x+\pi \right)=-\mathrm{cos}x$ then it's the same as
$\underset{x\to 0}{lim}\frac{1-\mathrm{cos}x}{1-\mathrm{cos}3x}$
As
$\mathrm{cos}3x=4{\mathrm{cos}}^{3}x-3\mathrm{cos}x$
it's the same as
$\underset{x\to 0}{lim}\frac{1-\mathrm{cos}x}{1+3\mathrm{cos}x-4{\mathrm{cos}}^{3}x}=\underset{x\to 0}{lim}\frac{1}{1+4\mathrm{cos}x+4{\mathrm{cos}}^{2}x}$
etc.

Boehm98wy

Enforcing $t=\pi -x$ and given the classical limit,
$\underset{x\to 0}{lim}\frac{1-\mathrm{cos}x}{{x}^{2}}=\frac{1}{2}$
we have
$\underset{t\to \pi }{lim}\frac{1+\mathrm{cos}t}{1+\mathrm{cos}3t}$
$=\underset{x\to 0}{lim}\frac{1-\mathrm{cos}x}{1-\mathrm{cos}3x}=\frac{1}{9}\underset{x\to 0}{lim}\frac{1-\mathrm{cos}x}{{x}^{2}}\frac{{\left(3x\right)}^{2}}{1-\mathrm{cos}3x}$
$=\frac{1}{9}\underset{x\to 0}{lim}\frac{1-\mathrm{cos}x}{{x}^{2}}\underset{x\to 0}{lim}\frac{{\left(3x\right)}^{2}}{1-\mathrm{cos}3x}$
$=\frac{1}{9}\underset{x\to 0}{lim}\frac{1-\mathrm{cos}x}{{x}^{2}}\underset{h\to 0}{lim}\frac{{h}^{2}}{1-\mathrm{cos}h}$
$=\frac{1}{9}$

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