Question

Find $k$ such that function is continuous at $x=0$

$f\left(x\right)=\frac{x^3-3x}{sinx}$ for $x\ne 0$

$=k$ for $x=0$

Answer 1

For function to be continuous $x=0$, limit of function must be finite at this point.

$\underset{x\to 0}{lim}\frac{x^3-3x}{sinx}=\underset{x\to 0}{lim}\frac{3x^2-3}{cosx}=\frac{0-3}{1}=-3$.

Hence $k=-3$.