Question

$\underset{x\to 0}{lim}\left(\frac{\sin x-x}{x}\right)\left(\sin \frac{1}{x}\right)$ is:

• A. does not exist
• B. is equal to 0
• C. is equal to 1
• D. exists and different from 0 and 1

Answer 1

Answer: (B)

is equal to 0

$sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}....\left(1\right)$
$sinx-x=-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}$
$\frac{sinx-x}{x}=-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}....=0\left(asxfoundto0\right)$
sin$\left(\frac{1}{x}\right)$ is an oscillatory function means same finite no.
$0\times finiteno.=0$