Question

If $\underset{x\to m}{lim}\frac{x^3-m^3}{x-m}=3$ then find the number of possible values of $m$.

• A. $0$
• B. $2$
• C. $1$
• D. $3$

Answer 1

The correct option is B $2$

$\underset{x\to m}{lim}\frac{x^3-m^3}{x-m}=3$
We know that, $\underset{x\to a}{lim}\frac{x^n-a^n}{x-a}=n{a}^{n-1}$
$\Rightarrow 3m^{3-1}=3$
$\Rightarrow 3m^2=3$
$\Rightarrow m^2=1$
$\Rightarrow m=\pm 1$
Therefore, there are $2$ possible values of $m$.