proving that limx→π1+cos⁡x1+cos⁡3x=19 without L'Hôpital

As cos⁡(x+π)=−cos⁡x then it's the same aslimx→01−cos⁡x1−cos⁡3xAscos⁡3x=4cos3x−3cos⁡xit's the same aslimx→01−cos⁡x1+3cos⁡x−4cos3x=limx→011+4cos⁡x+4cos2xetc.

Enforcing t=π−x and given the classical limit,limx→01−cos⁡xx2=12we havelimt→π1+cos⁡t1+cos⁡3t=limx→01−cos⁡x1−cos⁡3x=19limx→01−cos⁡xx2(3x)21−cos⁡3x=19limx→01−cos⁡xx2limx→0(3x)21−cos⁡3x=19limx→01−cos⁡xx2limh→0h21−cos⁡h=19

As cos⁡(x+π)=−cos⁡x then it's the same aslimx→01−cos⁡x1−cos⁡3xAscos⁡3x=4cos3x−3cos⁡xit's the same aslimx→01−cos⁡x1+3cos⁡x−4cos3x=limx→011+4cos⁡x+4cos2xetc.

Enforcing t=π−x and given the classical limit,limx→01−cos⁡xx2=12we havelimt→π1+cos⁡t1+cos⁡3t=limx→01−cos⁡x1−cos⁡3x=19limx→01−cos⁡xx2(3x)21−cos⁡3x=19limx→01−cos⁡xx2limx→0(3x)21−cos⁡3x=19limx→01−cos⁡xx2limh→0h21−cos⁡h=19

Explained: "proving that limx→π1+cos⁡x1+cos⁡3x=19 without L'Hôpital"

michiiiiiakqm

Answered question

2022-03-28

proving that

lim_{x \to π} \frac{1 + \cos x}{1 + \cos 3 x} = \frac{1}{9}

without L'Hôpital

Answer & Explanation

Avery Maxwell

Beginner2022-03-29Added 13 answers

\cos (x + π) = - \cos x

then it's the same as

lim_{x \to 0} \frac{1 - \cos x}{1 - \cos 3 x}

\cos 3 x = 4 \cos^{3} x - 3 \cos x

it's the same as

lim_{x \to 0} \frac{1 - \cos x}{1 + 3 \cos x - 4 \cos^{3} x} = lim_{x \to 0} \frac{1}{1 + 4 \cos x + 4 \cos^{2} x}

etc.

Boehm98wy

Beginner2022-03-30Added 18 answers

Enforcing

t = π - x

and given the classical limit,

lim_{x \to 0} \frac{1 - \cos x}{x^{2}} = \frac{1}{2}

we have

lim_{t \to π} \frac{1 + \cos t}{1 + \cos 3 t}

= lim_{x \to 0} \frac{1 - \cos x}{1 - \cos 3 x} = \frac{1}{9} lim_{x \to 0} \frac{1 - \cos x}{x^{2}} \frac{{(3 x)}^{2}}{1 - \cos 3 x}

= \frac{1}{9} lim_{x \to 0} \frac{1 - \cos x}{x^{2}} lim_{x \to 0} \frac{{(3 x)}^{2}}{1 - \cos 3 x}

= \frac{1}{9} lim_{x \to 0} \frac{1 - \cos x}{x^{2}} lim_{h \to 0} \frac{h^{2}}{1 - \cos h}

= \frac{1}{9}

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