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  • Find the value of:   <img alt="\mathrm{\lim _{x \rightarrow 0}\left\{\frac{2}{x^3}(\tan x-\sin x)\right\}^{2 / x^2}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cl

Find the value of:   \mathrm{\lim _{x \rightarrow 0}\left\{\frac{2}{x^3}(\tan x-\sin x)\right\}^{2 / x^2}}

Option: 1

\mathrm{\sqrt{e}}


Option: 2

\mathrm{\sqrt{2e}}


Option: 3

0


Option: 4

1


Answers (1)

best_answer

Hint:

Determine first the limit of

                          \mathrm{\ln \left(\frac{2}{x^3}(\tan x-\sin x)\right)^{\frac{2}{x^2}}=\frac{2}{x^2} \ln \left(\frac{2}{x^3}(\tan x-\sin x)\right)}

For that, you need to expand \mathrm{\tan x-\sin x} at order 5:

\mathrm{\tan x-\sin x=x+\frac{x^3}{3}+\frac{2}{15} x^5+o\left(x^5\right)-\left(x-\frac{x^3}{6}+\frac{x^5}{120}+o\left(x^5\right)\right)=\frac{x^3}{2}+\frac{x^5}{8}+o\left(x^5\right),}

                          so               \mathrm{\ln \left(\frac{2}{x^3}(\tan x-\sin x)\right)=\ln \left(1+\frac{x^2}{4}+o\left(x^2\right)\right)=\frac{x^2}{4}+o\left(x^2\right)}

                                                 and finally    \mathrm{\frac{2}{x^2} \ln \left(\frac{2}{x^3}(\tan x-\sin x)\right)=\frac{1}{2}+o(1) \rightarrow \frac{1}{2}}

As a conclusion , the limit of given expression is \mathrm{\sqrt{e}}  .

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