(IME-CG) Limites

Prova do IME vale Séries de Maclaurin? Veja:

[latex]\\\mathrm{Por\ S\acute{e}ries\ de\ Maclaurin:\left\{\begin{matrix} \mathrm{sin(x)\approx x}\\ \mathrm{sin(2x)\approx 2x}\\ \mathrm{sin(3x)\approx 3x} \end{matrix}\right.}\\\\ \mathrm{\lim_{x\to 0}\frac{\sqrt[3]{\mathrm{sin(x)sin(2x)sin(3x)}}}{|x|}=\lim_{x\to 0}\frac{\sqrt[3]{\mathrm{6x^3}}}{|x|}}\\\\ \mathrm{Dado\ que\ |x|=\left\{\begin{matrix} \mathrm{x,se\ x\geq 0}\\ \mathrm{-x,se\ x<0} \end{matrix}\right.}\\\\ \mathrm{\lim_{x\to 0^-}\frac{\sqrt[3]{\mathrm{sin(x)sin(2x)sin(3x)}}}{|x|}=\lim_{x\to 0^-}\frac{\sqrt[3]{\mathrm{6x^3}}}{|x|}=\frac{x\sqrt[3]{\mathrm{-6}}}{x}=\sqrt[3]{\mathrm{-6}}}\\\\ \mathrm{\lim_{x\to 0^+}\frac{\sqrt[3]{\mathrm{sin(x)sin(2x)sin(3x)}}}{|x|}=\lim_{x\to 0^+}\frac{\sqrt[3]{\mathrm{6x^3}}}{|x|}=\frac{x\sqrt[3]{\mathrm{6}}}{x}=\sqrt[3]{\mathrm{6}}}\\\\\ \mathrm{\lim_{x\to 0^-}f(x)\neq \lim_{x\to 0^+}f(x)\ \therefore\ \lim_{x\to 0}f(x)= \nexists }[/latex]

Só por desencargo de consciência, vou resolver usando um jeito mais tradicional. Veja se você concorda.

[latex]\mathrm{\lim_{x\to0}\frac{\sqrt[3]{\mathrm{sin(x)sin(2x)sin(3x)}}}{|x|}=\sqrt[3]{\mathrm{6}}\lim_{x\to 0}\sqrt[3]{\mathrm{\frac{sin(x)sin(2x)sin(3x)}{6|x|^3}}}}[/latex]

[latex]\mathrm{\lim_{x\to0}\frac{\sqrt[3]{\mathrm{sin(x)sin(2x)sin(3x)}}}{|x|}=\sqrt[3]{\mathrm{6}}\sqrt[3]{\mathrm{\lim_{x\to 0}\frac{sin(x)}{|x|}\lim_{x\to 0}\frac{sin(2x)}{2|x|}\lim_{x\to 0}\frac{sin(3x)}{3|x|}}}}[/latex]

[latex]\mathrm{Dado\ que\ |x|=\left\{\begin{matrix}
\mathrm{x,se\ x\geq 0}\\
\mathrm{-x,se\ x<0}
\end{matrix}\right.}[/latex]

[latex]\mathrm{\lim_{x\to0^-}\frac{\sqrt[3]{\mathrm{sin(x)sin(2x)sin(3x)}}}{|x|}=\sqrt[3]{\mathrm{6}}\sqrt[3]{\mathrm{-\lim_{x\to0^-}\frac{sin(x)}{x}\lim_{x\to0^-}\frac{sin(2x)}{2x}\lim_{x\to0^-}\frac{sin(3x)}{3x}}}=\sqrt[3]{\mathrm{-6}}}[/latex]

[latex]\mathrm{\lim_{x\to0^+}\frac{\sqrt[3]{\mathrm{sin(x)sin(2x)sin(3x)}}}{|x|}=\sqrt[3]{\mathrm{6}}\sqrt[3]{\mathrm{\lim_{x\to0^+}\frac{sin(x)}{x}\lim_{x\to0^+}\frac{sin(2x)}{2x}\lim_{x\to0^+}\frac{sin(3x)}{3x}}}=\sqrt[3]{\mathrm{6}}}[/latex]

[latex]\mathrm{\lim_{x\to 0^-}f(x)\neq \lim_{x\to 0^+}f(x) \ \therefore\ \lim_{x\to 0}f(x)=\nexists}[/latex]