Question

If $\underset{x\to -2}{lim}\frac{x^p+2^p}{x+2}=80$ (where $p$ is an odd number), then $p$ can be _____

• A. $3$
• B. $5$
• C. $7$
• D. $9$

Answer 1

The correct option is B $5$

$\underset{x⟶-2}{lim}\frac{x^p+2^p}{x+2}=80$ ........... [$P$ is an odd number]
Put $x=-2$, we get $\frac{0}{0}$form.
Apply L-Hospital's rule, we have
$\underset{x⟶-2}{lim}\frac{{px}^{p-1}+0}{1+0}=80$

$\Rightarrow p{(-2)}^{p-1}=80$
$\{p$ is odd $\Rightarrow (p-1)$ is even
Therefore, ${(-2)}^{p-1}={\left(2\right)}^{p-1}\}$

$\Rightarrow p{\left(2\right)}^{p-1}=5\times 16$
$\Rightarrow p{\left(2\right)}^{p-1}=5\times 2^{5-1}$
$\Rightarrow p=5$