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Computing $\lim_{x\to{0+}}\frac{\tan(x)-x}{x^3}$ without L'Hopital

Say $\lim_{x\to{0+}}\frac{\tan(x)-x}{x^3} = L$

For $L$: $$L=\lim_{x\to0}\frac{\tan x-x}{x^3}\\ L=\lim_{x\to0}\frac{\tan 2x-2x}{8x^3}\\ 4L=\lim_{x\to0}\frac{\frac12\tan2x-x}{x^3}\\ 3L=\lim_{x\to0}\frac{\frac12\tan{2x}-\tan x}{x^3}\\ =\lim_{x\to0}\frac{\tan x}x\frac{\frac1{1-\tan^2x}-1}{x^2}\\ =\lim_{x\to0}\frac{(\tan x)^3}{x^3}=1\\ \large L=\frac13$$

I found that in another Q, can someone tell me why

$$L=\lim_{x\to0}\frac{\tan x-x}{x^3}=\lim_{x\to0}\frac{\tan 2x-2x}{8x^3}$$

N. F. Taussig
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Juju9708
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3 Answers3

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If $x = 2y$ then $y\rightarrow 0$ when $x\rightarrow 0$, so $\lim_{x\rightarrow0} f(x) = \lim_{y\rightarrow 0} f(2y)$.

Alex Jones
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Use the McLaurian series of $\tan x$ which is simply $$\tan x=x+\frac {x^3}{3}+ O(x^5)$$

Hence $$\lim_{x\to 0^+} \frac {\tan x-x}{x^3}=\lim_{x\to 0^+} \frac {x+x^3/3-x}{x^3}=\lim_{x\to 0^+} \frac 13=\frac 13$$

Similarly for second one just replace $x$ with $2x$

Rohan Shinde
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    This is a good answer to the question, but I want to add that usually restrictions against using L'Hopital's rule come from a point in learning Calculus during which you don't know Taylor/McLaurin series. So, for anyone who hasn't learned to show the first line of this proof, they should not use it. – Alex Jones Apr 30 '18 at 06:58
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Let $x=\arctan y.$ We have $y\to 0^+$ as $x\to 0^+.$

For $x\in (0,\pi /2)$ we have $\frac {\tan x-x}{x^3}=$ $\frac {y-\arctan y}{y^3}\frac {y^3}{x^3}=$ $\frac {(y-\arctan y)}{y^3}(\frac {\tan x}{x})^3.$

We have $\lim_{x\to 0^+}(\frac {\tan x }{x})^3=1.$

Consider $g(y)=y-\arctan y.$

We have $g'(y)-y^2=$ $(1-\frac {1}{1+y^2})-y^2=$ $\frac {-y^4}{1+y^2}\in [-y^4,0].$

Therefore $g(y)=g(y)-g(0)=$ $\int_0^y g'(t)dt=$ $y^3/3-\delta \cdot y^5/5$ for some $\delta \in [0,1].$

DanielWainfleet
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