Question

Find the positive integer $n$ so that $\underset{x\to 3}{\lim }$ $\frac{x^n-3^n}{x-3}=108$.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (D)

We have, $\underset{x\to 3}{\lim }$ $\frac{x^n-3^n}{x-3}$
$=n(3{)}^{n-1}$
Therefore, $n(3{)}^{n-1}=108$
$=4(27)=4(3{)}^{4-1}$
On comparing, we get $n=4$