Question

The function $f\left(x\right)=\frac{3\sin x-\sin 3x}{x^3}$ for $x\ne 0$;
$f\left(0\right)=1$ at $x=0$

• A. Continuous
• B. Discontinuous
• C. Not determined
• D. None

Answer 1

Answer: (B)

Discontinuous

Now,

$\underset{x\to 0}{lim}f\left(x\right)$

$=\underset{x\to 0}{lim}\frac{3\sin x-\sin 3x}{x^3}$

$=\underset{x\to 0}{lim}\frac{4{\sin }^{3}x}{x^3}$ [ Since $\sin 3x=3\sin x-4{\sin }^{3}x$]

$=4{\left(\underset{x\to 0}{lim}\frac{\sin x}{x}\right)}^3$

$=4\times 1^3$

$=4$.

But $\underset{x\to 0}{lim}f\left(x\right)\ne f\left(0\right)$. [ Since $f\left(0\right)=1$]

So the given function is discontinuous at $x=0$.