Question

Limita când x->0 din (tgx-x)/x la puterea a 3 v

Answer 1

$\displaystyle \mathtt{ \lim_{x \to 0} \frac{tg~x-x}{x^3} \overset{\underset{\mathrm{l'H}}{}}{=} \lim_{x \to 0}\frac{(tg~x-x)'}{\left(x^3\right)'}= \lim_{x \to 0} \frac{(tg~x)'-x'}{3x^2}=}\\ \\ \mathtt{= \lim_{x \to 0} \frac{1+tg^2x-1}{3x^2}= \lim_{x \to 0} \frac{tg^2x}{3x^2}\overset{\underset{\mathrm{l'H}}{}}{=} \lim_{x \to 0} \frac{\left(tg^2x)'}{\left(3x^2\right)'}= \lim_{x \to 0} \frac{(tg~x\cdot tg~x)'}{6x} =}$
$\displaystyle \mathtt{= \lim_{x \to 0} \frac{(tg~x)' \cdot tg~x+tg~x\cdot(tg~x)'}{6x}= \lim_{x \to 0} \frac{ \frac{1}{cos^2x}\cdot tg~x+tg~x \cdot \frac{1}{cos^2x} }{6x}=}\\ \\ \mathtt{= \lim_{x \to 0}\frac{ \frac{2tg~x}{cos^2x} }{6x}= \lim_{x \to 0}\frac{2tg~x}{6xcos^2x}= \lim_{x \to 0} \frac{tg~x}{3xcos^2x}\overset{\underset{\mathrm{l'H}}{}}{=} \lim_{x \to 0} \frac{(tg~x)'}{\left(3xcos^2x\right)'}=}$
$\displaystyle \mathtt{= \lim_{x \to 0} \frac{ \frac{1}{cos^2x} }{3\left(x'\cdot cos^2x+x \cdot \left(cos^2x\right)'\right)}= \lim_{x \to 0} \frac{ \frac{1}{cos^2x} }{3\left(cos^2x-xsin2x\right)}=}\\ \\ \mathtt{= \lim_{x \to 0} \frac{1}{3cos^2x\left(cos^2x-xsin2x\right)}= \frac{1}{3cos^20\left(cos^20-0\cdot sin0\right)}=}\\ \\ \mathtt{= \frac{1}{3 \cdot 1(1-0)}= \frac{1}{3 \cdot1} = \frac{1}{3} }\\ \\ \mathtt{ \lim_{x \to 0} \frac{tg~x-x}{x^3} = \frac{1}{3} }$