Question

Solve $\frac{2}{11}\underset{x\to 0}{lim}(\left[\frac{sinx}{x}\right]+\left[\frac{2sinx}{x}\right]+\left[\frac{3sinx}{x}\right]+...+\left[\frac{11sinx}{x}\right])$

• A. $8$
• B. $9$
• C. $10$
• D. $12$

Answer 1

Answer: (D)

$12$

$\frac{2}{11}\{\begin{array}{c}\underset{x\to 0}{lim}\left[\frac{sinx}{x}\right]+2\underset{x\to 0}{lim}\frac{sinx}{x}\\ ....\underset{x\to 0}{lim}\frac{sinx}{x}\end{array}$
$=\frac{2}{11}[1+2+3+4...11]$
$\frac{2}{11}\frac{11(11+1)}{2}$
$=12$