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  • Evaluate the given limit: <img alt="\mathrm{\lim _{x \rightarrow 0} \frac{\ln \left(x+\sqrt{1+x^2}\right)-x}{\tan ^3(x)} .}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5

Evaluate the given limit: \mathrm{\lim _{x \rightarrow 0} \frac{\ln \left(x+\sqrt{1+x^2}\right)-x}{\tan ^3(x)} .}

Option: 1

-\frac{1}{6}


Option: 2

\frac{2}{3}


Option: 3

0


Option: 4

1


Answers (1)

best_answer

i will use the maclaurin expansion for \mathrm{\sqrt{1+x}, \ln (1+x), \tan (x)}

\mathrm{ \begin{aligned} \ln \left[x+\left(1+x^2\right)^{1 / 2}\right] & =\ln \left[x+1+\frac{1}{2} x^2-\frac{1}{8} x^4+\cdots\right] \\ & =\ln \left(1+x+\frac{1}{2} x^2-\frac{1}{8} x^4+\cdots\right) \\ & =\left(x+\frac{1}{2} x^2-\frac{1}{8} x^4+\cdots\right)-\frac{1}{2}\left\{x+\frac{1}{2} x^2-\frac{1}{8} x^4+\cdots\right\}^2 \\ & +\frac{1}{3}\left\{x+\frac{1}{2} x^2-\frac{1}{8} x^4+\cdots\right\}^3 \cdots \\ & =\left(x+\frac{1}{2} x^2-\frac{1}{8} x^4+\cdots\right)-\frac{x^2}{2}(1+x+\cdots)+\frac{1}{3}\left(x^3+\cdots\right)+\cdots \\ & =x-\frac{1}{2} x^3+\frac{1}{3} x^3+\cdots \\ & =x-\frac{1}{6} x^3+\cdots \end{aligned} }the expansion for 

                                                                  \mathrm{\tan x=x+\cdots}

putting the two together 

                                                 \mathrm{\lim _{x \rightarrow 0} \frac{\ln \left[x+\left(1+x^2\right)^{1 / 2}\right]-x}{\tan ^3 x}=-\frac{1}{6}}

Posted by

Rishi

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