# Limits by Taylor series

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When we want to evaluate a limit of a function, it is sometimes useful to know the Taylor series of the function itself, or an approximation if it is seasonable.

Prove

$\displaystyle \lim_{x\to0} \dfrac{\tan x - x}{x^3} = \dfrac13.$

Knowing the Taylor series of $\tan x$ centered at $x=0$ is $x+ \frac13 x^3 + O\big(x^5\big)$, we have

$\dfrac{\tan x - x}{x^3} = \dfrac{\frac13 x^3 + O\big(x^5\big) }{x^3} ={\dfrac13 x^2 + O\big(x^2\big)}.$

Hence,

$\displaystyle \lim_{x\to0} \dfrac{\tan x - x}{x^3} = \lim_{x\to0} \left( \dfrac13 x^2 + O\big(x^2\big)\right) = \dfrac13.\ _\square$

## See Also

**Cite as:**Limits by Taylor series.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/limits-by-taylor-series/