Question

The value of $\underset{x \rightarrow 0}{\text{lim}}\frac{x^{6 0 0 0} - \left(sin \, x\right)^{6 0 0 0}}{x^{2} \left(sin x\right)^{6 0 0 0}}$ is

• A. $1000$
• B. $100$
• C. $1100$
• D. $1010$

Answer 1

Answer: (A)

$\underset{x \rightarrow 0}{lim} \frac{x^{6000} - \left(sin x\right)^{6000}}{x^{2} \frac{\left(sin x\right)^{6000}}{x^{6000}} . x^{6000}}\left(\frac{0}{0}\right)$
$\underset{x \rightarrow 0}{\text{lim}}\frac{x^{6 0 0 0} - \left(sin x\right)^{6 0 0 0}}{x^{6 0 0 2}}$
$= -\underset{x \rightarrow 0}{\text{lim}}\frac{\left(sin x\right)^{6 0 0 0} - x^{6 0 0 0}}{x^{6 0 0 2}}$
$=-\underset{x \rightarrow 0}{\text{lim}}\frac{\left(\frac{sin x}{x}\right)^{6 0 0 0} - 1}{x^{2}}$
$=-\underset{x \rightarrow 0}{\text{lim}}\frac{6 0 0 0 \left(\frac{sin x}{x}\right)^{5 9 9 9} \left(\frac{x cos x - sin x}{x^{2}}\right)}{2 x}$ (By L-Hospital Rule)
$=-\frac{6 0 0 0}{2}\underset{x \rightarrow 0}{\text{lim}}\frac{x cos x - sin x}{x^{3}}$

$=-3000 \, \underset{x \rightarrow 0}{lim}\frac{cos x - x \, sin x - cos x}{3 x^{2}} \, $
$=1000$