Tuesday, June 27, 2017

calculus - Computing $lim_{xto{0+}}frac{tan(x)-x}{x^3}$ without L'Hôpital's rule.






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



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


Answer



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



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