Question

Let $f\left(x\right)=\left[\frac{\sin X}{X}\right]+\left[\frac{2\sin 2X}{X}\right]+....+\left[\frac{10\sin 10X}{X}\right]$ (where [.] denotes G.I.F).The value of ${lim}_{X\to 0}f\left(x\right)$ equals

• A. $55$
• B. $164$
• C. $165$
• D. $375$

Answer 1

The correct option is D $375$

By graph we can say that,
${lim}_{X\to 0}\left[\frac{\sin X}{X}\right]=0=1^2-1$ similarly
$\frac{2\sin 2X}{X}=\frac{4\sin 2X}{2X}$
${lim}_{X\to 0}\frac{4\sin 2X}{2X}=4$
${lim}_{X\to 0}\frac{4\sin 2X}{2X}=3=2^2-1$
${lim}_{X\to 0}\frac{n\sin nX}{X}=n^2-1$
$\left[\frac{\sin X}{X}\right]+\left[\frac{2\sin 2X}{X}\right]+....+\left[\frac{10\sin 10X}{X}\right]$ for ${lim}_{X\to 0}$
$1^2-1+2^2-1+...{.10}^2-1$
$1^2+2^2+3^2+...{.10}^2-10$
$385-10=375$