问题及解答

求极限

\[
\lim_{x\rightarrow 0}\frac{6e^{-x^2}\sin x-x(6-7x^2)}{3\ln\frac{1+x}{1-x}-2x(3+x^2)}
\]

原极限等于

\[
\begin{split}
&\lim_{x\rightarrow 0}\frac{6e^{-x^2}\sin x-6x+7x^3}{3\ln\frac{1+x}{1-x}-6x-2x^3}\\
\stackrel{\frac{0}{0}}{=}&\lim_{x\rightarrow 0}\frac{6e^{-x^2}(-2x)\sin x+6e^{-x^2}\cos x-6+21x^2}{3\cdot\frac{1-x}{1+x}\cdot\frac{1\cdot(1-x)-(1+x)\cdot(-1)}{(1-x)^2}-6-6x^2}\\
=&\lim_{x\rightarrow 0}\frac{2e^{-x^2}(-2x\sin x+\cos x)-2+7x^2}{\frac{2}{1-x^2}-2(1+x^2)}\\
=&\lim_{x\rightarrow 0}\frac{e^{-x^2}(-2x\sin x+\cos x)-1+\frac{7}{2}x^2}{\frac{x^4}{1-x^2}}\\
=&\lim_{x\rightarrow 0}\frac{(-2x\sin x+\cos x)+e^{x^2}(\frac{7}{2}x^2-1)}{x^4e^{x^2}}\\
\stackrel{\frac{0}{0}}{=}&\lim_{x\rightarrow 0}\frac{-2\sin x-2x\cos x-\sin x+e^{x^2}\cdot(2x)(\frac{7}{2}x^2-1)+e^{x^2}\cdot 7x}{4x^3e^{x^2}+x^4 e^{x^2}\cdot 2x}\\
=&\lim_{x\rightarrow 0}\frac{-3\frac{\sin x}{x}-2\cos x+e^{x^2}(7x^2+5)}{e^{x^2}(4x^2+2x^4)}\\
=&\lim_{x\rightarrow 0}\frac{e^{x^2}\cdot 7x^2}{e^{x^2}(4x^2+2x^4)}+\lim_{x\rightarrow 0}\frac{5e^{x^2}-3\frac{\sin x}{x}-2\cos x}{e^{x^2}\cdot 2x^2(2+x^2)}\\
=&\frac{7}{4}+\lim_{x\rightarrow 0}\frac{5xe^{x^2}-3\sin x-2x\cos x}{4x^3}\\
=&\frac{7}{4}+\lim_{x\rightarrow 0}\frac{5(e^{x^2}+xe^{x^2}\cdot 2x)-3\cos x-2(\cos x-x\sin x)}{12x^2}\\
=&\frac{7}{4}+\lim_{x\rightarrow 0}\frac{5e^{x^2}(1+2x^2)-5\cos x+2x\sin x}{12x^2}\\
=&\frac{7}{4}+\lim_{x\rightarrow 0}\frac{5e^{x^2}\cdot 2x(1+2x^2)+5e^{x^2}\cdot 4x+5\sin x+2\sin x+2x\cos x}{24x}\\
=&\frac{7}{4}+\lim_{x\rightarrow 0}\frac{5e^{x^2}(4x^3+6x)+7\sin x+2x\cos x}{24x}\\
=&\frac{7}{4}+\lim_{x\rightarrow 0}\frac{5e^{x^2}\cdot 2x(4x^3+6x)+5e^{x^2}(12x^2+6)+7\cos x+2\cos x-2x\sin x}{24}\\
=&\frac{7}{4}+\frac{39}{24}\\
=&\frac{27}{8}.
\end{split}
\]