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Evaluate \mathrm{\lim _{x \rightarrow 0}\frac{\ln \left[\frac{(1-3 x)(1+x)^{3}}{(1+3 x)(1-x)^{3}}\right]}{x^{3}}}

Option: 1

-16


Option: 2

0


Option: 3

1


Option: 4

2


Answers (1)

best_answer

\mathrm{\lim _{x \rightarrow 0} \frac{\ln \left[\frac{(1-3 x)(1+x)^{3}}{(1+3 x)(1-x)^{3}}\right]}{x^{3}}=\lim _{x \rightarrow 0} \frac{\ln \left((1-3 x)(1+x)^{3}\right)-\ln \left((1+3 x)(1-x)^{3}\right)}{x^{3}}}
                                           \mathrm{=\lim _{x \rightarrow 0} \frac{\ln \left(1-6 x^{2}-8 x^{3}-3 x^{4}\right)-\ln \left(1-6 x^{2}+8 x^{3}-3 x^{4}\right)}{x^{3}}}
                                           \mathrm{=\lim _{x \rightarrow 0} \frac{\ln \left(1-6 x^{2}-8 x^{3}\right)-\ln \left(1-6 x^{2}+8 x^{3}\right)}{x^{3}}}
                                           \mathrm{ =\lim _{h \rightarrow 0} \frac{\ln \left(1-6 h^{2 / 3}-8 h\right)-\ln \left(1-6 h^{2 / 3}+8 h\right)}{h}}
                                           \mathrm{ =\left.\left\{\ln \left(1-6 x^{2 / 3}-8 x\right)-\ln \left(1-6 x^{2 / 3}+8 x\right)\right\}^{\prime}\right|_{x=0}}
                                           \mathrm{ =\left.\left(\frac{-\frac{4}{\sqrt[3]{x}}-8}{1-6 x^{2 / 3}-8 x}-\frac{-\frac{4}{\sqrt[3]{x}}+8}{1-6 x^{2 / 3}+8 x}\right)\right|_{x=0} }
                                           \mathrm{ =\left.\frac{-16+32 x^{2 / 3}}{\left(1-6 x^{2 / 3}-8 x\right)\left(1-6 x^{2 / 3}+8 x\right)}\right|_{x=0} }
                                           \mathrm{ =-16 }

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manish

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