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  • Evaluate    <img alt="\mathrm{L_{\mathrm{5}}=\lim _{x \rightarrow 0} \frac{\sin ^{-1} x-x}{x^3}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BL_%7B%5Cmathrm%7B5%

Evaluate    \mathrm{L_{\mathrm{5}}=\lim _{x \rightarrow 0} \frac{\sin ^{-1} x-x}{x^3}}

Option: 1

\frac{1}{6}


Option: 2

\frac{3}{5}


Option: 3

\frac{2}{3}


Option: 4

\frac{1}{2}


Answers (1)

best_answer

        \mathrm{\begin{gathered} L_5=\lim _{x \rightarrow 0} \frac{\sin ^{-1} x-x}{x^3} \\ 8 L_5=\lim _{x \rightarrow 0} \frac{\sin ^{-1} 2 x-2 x}{x^3} \\ 4 L_5=\lim _{x \rightarrow 0} \frac{\frac{1}{2} \sin ^{-1} 2 x-x}{x^3} \\ 3 L_5=\lim _{x \rightarrow 0} \frac{\frac{1}{2} \sin ^{-1} 2 x-\sin ^{-1} x}{x^3} \\ 6 L_5=\lim _{x \rightarrow 0} \frac{\sin ^{-1} 2 x-2 \sin ^{-1} x}{x^3} \end{gathered}}

\mathrm{\begin{gathered} 6 L_5=\lim _{x \rightarrow 0} \frac{\sin ^{-1}\left(-4 x^3-2 \sqrt{1-4 x^2} \sqrt{1-x^2} x+2 x\right)}{x^3} \\ 6 L_5=\lim _{x \rightarrow 0} \frac{-4 x^3+2 x\left(1-\sqrt{1-4 x^2} \sqrt{1-x^2}\right)}{x^3} \\ 6 L_5=\lim _{x \rightarrow 0}-4+2 \frac{\left(1-\sqrt{1-5 x^2+4 x^4}\right)}{x^2} \end{gathered}}

`            \mathrm{6 L_5=\lim _{x \rightarrow 0}-4+2 \frac{\left(1-\sqrt{1-5 x^2+4 x^4}\right)}{x^2}}

Since you would consider binolmial theorem as series expansion, if not well and good , if i'll do: Now let \mathrm{\sqrt{1-5 x^2+4 x^4}=\sum a t x^k} squaring both sides,

\mathrm{\begin{aligned} 1-5 x^2+4 x^4= & a_0^2+2 a_0 a_1 x+\left(2 a_0 a_2+a_1^2\right) x^2+\left(2 a_0 a_3+a_1 a_2\right) x^3 \\ & +\left(2 a_0 a_4+2 a_1 a_3+a_2^2\right) x^4+\ldots \end{aligned}}

now taking positive branch :

\mathrm{a_0=1, a_1=0, a_2=-5 / 2, a_3=0, a_4=-9 / 8, \ldots}

So: \mathrm{\begin{gathered} 6 L_5=\lim _{x \rightarrow 0}-4+2 \frac{\left(1-\left(1-5 x^2 / 2-9 x^4 / 8 \ldots\right)\right)}{x^2} \\ L_5=\frac{1}{6} \end{gathered}}

 

Posted by

Rakesh

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