Question

$\underset{x\to 0}{lim}\left(\frac{1}{{\sin }^{2}x}-\frac{1}{\sin h^{2}x}\right)=$

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. 0
• D. 1

Answer 1

Answer: (B)

$\frac{2}{3}$

• using expansion of series of sin x and sin h x

$\sin x=x-\frac{1}{6}{x}^3$ - $\frac{1}{120}{x}^5$....

$\sinh x=x+\frac{1}{6}{x}^3$ + $\frac{1}{120}{x}^5$....

${\sin }^{2}x$ = $x^2$ - $\frac{1}{3}{x}^4+\frac{32x^6}{6!}$

${\sinh }^{2}x$ = $x^2$ + $\frac{1}{3}{x}^4+\frac{13x^6}{360}$

limit x$\to$0 ($\frac{1}{{\sin }^{2}x}$-$\frac{1}{{\sinh }^{2}x}$)=

limit x$\to$0 $\frac{{\sin }^{2}hx-{\sin }^{2}x}{{\sin }^{2}x{\sin }^{2}hx}$

=limit x$\to$0 $\frac{x^2+\frac{x^4}{3}+\frac{13x^6}{360}-x^2+\frac{x^4}{3}-\frac{32x^6}{6!}}{(x^2+\frac{x^4}{3}+\frac{13x^6}{360})(x^2-\frac{x^4}{3}+\frac{32x^6}{6!})}$

cancelling $x^2$

and dividing by $x^4$ we get

limit x$\to$0$\frac{\frac{2}{3}+\frac{13x^2}{360}+\frac{32x^2}{6!}}{(1+\frac{x^4}{3}+\frac{13x^6}{360})(1-\frac{x^4}{3}+\frac{32x^6}{6!})}$

now putting value of limit

we get the desired limit (solution ) of $\frac{2}{3}$