Question

The value of $\underset{x\Rightarrow \infty }{lim}(\left[\frac{100x}{sinx}\right]+\left[\frac{99sinx}{x}\right])$, where [.] denotes the greatest integer function, is

• A. 197
• B. 198
• C. 199
• D. Does not exist

Answer 1

The correct option is B 198

\(For~x>0 \\
sin x<x \\
\Rightarrow \frac{sin x}{x}<1 \\
\therefore \left [ \frac{sin x}{x} \right ]=98 \\
and~\frac{x}{sin x}>1 \\
\therefore \frac{100 x}{sin x}=100 \\
\Rightarrow \left [ \frac{100 x}{sin x} \right ]=100 \\
\therefore \underset{x\rightarrow 0+}{lim} \left ( \left [ \frac{100 x}{sin x} \right ]+\left [ \frac{99 sin x}{x} \right ] \right ) \\
=100+98 \\
=98 \\
For~x<0 \\
sin x>x \\
For~x>0 \\
sin x<x \\
\Rightarrow \frac{sin x}{x}<1 \\
\therefore \left [ \frac{99 sin x}{x} \right ]<99 \\
\Rightarrow \left [ \frac{99 sin x}{x} \right ]=99 \\
and~\frac{x}{sin x}>1 \\
\therefore \frac{100 x}{sin x}>100 \\
\Rightarrow \left [ \frac{100 x}{sin x} \right ]=100 \\
\underset{x\rightarrow 0-}{lim}\left ( \left [ \frac{100 x}{sin x} \right ]+\left [ \frac{99 sin x}{x} \right ] \right ) \\
=100 + 98 = 198 \\
Hence,~\underset{x\rightarrow 0}{lim}\left ( \left [ \frac{100 x}{sin x} \right ]+\left [ \frac{99 sin x}{x} \right ] \right )=198