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

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

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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,

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

• Taylor series

• Maclaurin series

• L'Hôpital's rule

• Limits by logarithms

• Limits by factoring

• Indeterminate forms