Question

If $\underset{x\to a}{lim}\frac{x^9-a^9}{x-a}=\underset{x\to 5}{lim}(4+x)$ then the value of $|a|$ is

• A. $1$
• B. $-1$
• C. $-2$
• D. $2$

Answer 1

The correct option is A $1$

We have,

$\underset{x\to a}{lim}\frac{x^9-a^9}{x-a}={lim}_{x\to 5}(4+x)$

Apply L-hospital rule, we get

$\underset{x\to a}{lim}\frac{9x^8-0}{1-0}={lim}_{x\to 5}(4+x)$

$\underset{x\to a}{lim}9x^8={lim}_{x\to 5}(4+x)$

$\underset{x\to a}{lim}9x^8=4+5$

$limx\to a9x^8=9$

$\underset{x\to a}{lim}{x}^8=1$

$a^8=1$

$a=1$

$|a|=1$