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Compute    \mathrm{\lim _{x \rightarrow 0} \frac{(x+2) \cdot \ln (1+x)-2 x}{x^3}}

Option: 1

2/3


Option: 2

1/20


Option: 3

0


Option: 4

2/5


Answers (1)

best_answer

Note that

                     \mathrm{\int_0^x \frac{-t^3}{1+t} d t=\int_0^x\left(\frac{1}{1+t}-\left(1-t+t^2\right)\right) d t=\ln (1+x)-x+\frac{1}{2} x^2-\frac{1}{3} x^3}

and thus

                                          \mathrm{\lim _{x \rightarrow 0} \frac{2 \ln (1+x)+x^2-2 x}{x^3}=\frac{2}{3}+\lim _{x \rightarrow 0}\left(\frac{-2}{x^3} \int_0^x \frac{t^3}{1+t} d t\right)}

For  \mathrm{|x|<1} , we have

                     \mathrm{0 \leq\left|\frac{-2}{x^3} \int_0^x \frac{t^3}{1+t} d t\right| \leq \frac{2}{|x|^3} \int_0^{|x|} \frac{|x|^3}{1-|x|} d t=\frac{2|x|}{1-|x|} \rightarrow 0 \quad \text { as } x \rightarrow 0}

so by the squeeze Theorem, we wind up with 

                                                            \mathrm{\lim _{x \rightarrow 0} \frac{2 \ln (1+x)+x^2-2 x}{x^3}=\frac{2}{3}}

 

 

Posted by

jitender.kumar

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