Question

The sum of an infinite geometric series whose first term is the limit of the function $f\left(x\right)=\frac{\tan x-\sin x}{{\sin }^{3}x}$ as $x\to 0$ and whose common ratio is the limit of the function $g\left(x\right)=\frac{1-\sqrt{x}}{{\left({\cos }^{-1}x\right)}^2}\text{as}x\to 1$ is

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $\frac{2}{3}$

Answer 1

The correct option is D $\frac{2}{3}$

$a=\underset{x\to 0}{lim}\frac{\tan x-\sin x}{{\sin }^{3}x}$

$=\underset{x\to 0}{lim}\frac{\frac{1}{\cos x}-1}{{\sin }^{2}x}$

$\underset{x\to 0}{lim}\frac{1-\cos x}{\cos x{\sin }^{2}x}$

$=\underset{x\to 0}{lim}\frac{(1-\cos x)}{\cos x(1+\cos x)(1-\cos x)}$

$=\underset{x\to 0}{lim}\frac{1}{\cos x(1+\cos x)}$

$=\frac{1}{1.\left(2\right)}=\frac{1}{2}$

$r=\underset{x\to 1}{lim}\frac{1-\sqrt{x}}{({\cos }^{-1}x{)}^2}$

$=\frac{-\frac{1}{2\sqrt{x}}}{2{\cos }^{-1}x\left(\frac{-1}{\sqrt{1-x^2}}\right)}$

$=\underset{x\to 1}{lim}=\frac{\sqrt{1-x^2}}{4\sqrt{x}{\cos }^{-1}x}$

$=\underset{x\to 1}{lim}\frac{\frac{11(-2x)}{4.2\sqrt{[1-x^2}}}{\frac{1}{2\sqrt{x}}{\cos }^{-1}x+\sqrt{x}\frac{(-1)}{\sqrt{1-x^2}}}$

$\underset{x\to 1}{lim}\frac{1}{4}\frac{\frac{-x}{\sqrt{1-x^2}}}{\frac{{\cos }^{-1}x}{2\sqrt{x}}-\frac{\sqrt{x}}{\sqrt{1-x^2}}}$

$\underset{x\to 1}{lim}\frac{1}{4}.\frac{-x}{\frac{\left({\cos }^{-1}x\right)\sqrt{1-x^2}}{2\sqrt{x}}-\sqrt{x}}$

$=\underset{x\to 1}{lim}\frac{1}{4}.\frac{-x.2\sqrt{x}}{\left({\cos }^{-1}x\right)\sqrt{1-x^2}-2x}$

$r=\frac{1}{4}\left(\frac{-1.2\sqrt{1}}{0-2.1}\right)=\frac{1}{4}$

$a=\frac{1}{2},r=\frac{1}{4}$

$s=\frac{a}{1-r}=\frac{1/2}{1-1/4}=\frac{\frac{2}{4}}{\frac{3}{4}}=\frac{2}{3}$