Question

lf the function defined by $f\left(x\right)=\frac{\sin 3(x-p)}{\sin 2(x-p)}$ for $x\ne p$ is continuous at $x=p$ then $f\left(p\right)=$

• A. $\frac{3}{2}$
• B. $\frac{2}{3}$
• C. $6$
• D. $\frac{1}{6}$

Answer 1

The correct option is A $\frac{3}{2}$

Since, it is given the function is continuous at $x=p,$we calculate

${lim}_{x\to p}\frac{sin3(x-p)}{sin2(x-p)}$

$=li{m}_{x\to p}\frac{sin3(x-p)}{3(x-p)}\times \frac{2(x-p)}{sin2(x-p)}\times \frac{3}{2}$

Using the property of limits,

$f\left(p\right)=1\times 1\times \frac{3}{2}$