Mathematics: Use Taylor series to evaluate limx→0(tan x − x)/x^3

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Use Taylor series to evaluate limx→0(tan x − x)/x^3

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idontknow113

09.08.2021

Mathematics

Recall that

tan(x) = sin(x)/cos(x)

and

sin(x) = x - x ³/6 + x ⁵/120 - x ⁷/5040 + …

cos(x) = 1 - x ²/2 + x ⁴/24 - x ⁶/720 + …

Truncate the series to three terms. Then

$\displaystyle \lim_{x\to0}\frac{\tan(x)-x}{x^3} = \lim_{x\to0}\frac{\frac{x-x^3/6+x^5/120}{1-x^2/2+x^4/24}-x}{x^3} \\\\ = \lim_{x\to0}\left(\frac{x-x^3/6+x^5/120}{x^3-x^5/2+x^7/24}-\frac1{x^2}\right) \\\\ = \lim_{x\to0}\left(\frac{1-x^2/6+x^4/120}{x^2-x^4/2+x^6/24}-\frac1{x^2}\right) \\\\ = \lim_{x\to0}\left(\frac{1-x^2/6+x^4/120}{x^2\left(1-x^2/2+x^4/24\right)}-\frac1{x^2}\right) \\\\ = \lim_{x\to0}\left(\frac{1-x^2/6+x^4/120}{x^2\left(1-x^2/2+x^4/24\right)}-\frac{1-x^2/2+x^4/24}{x^2\left(1-x^2/2+x^4/24\right)}\right) \\\\ = \lim_{x\to0}\frac{x^2/3-x^4/30}{x^2\left(1-x^2/2+x^4/24\right)} \\\\ = \lim_{x\to0}\frac{1/3-x^2/30}{1-x^2/2+x^4/24} = \boxed{\frac13}$

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Use Taylor series to evaluate
limx→0(tan x − x)/x^3

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