Question

Find the value of $\underset{x\to a}{lim}\frac{x^n-a^n}{x-a}$

• A.
• B.
• C.
• D. nx

Answer 1

The correct option is C

We have $\underset{x\to a}{lim}\frac{x^n-a^n}{x-a}$

If we substitude $x=a$ in the given limit, we find that numerator and

denominator of the limit becomes $\frac{0}{0}$.form

$\underset{x\to a}{lim}\frac{x^n-a^n}{x-a}$

Substituting $x=a+h$

$\underset{h\to 0}{lim}\frac{(a+h{)}^n-a^n}{a+h-a}$

$=\underset{h\to 0}{lim}\frac{a^n\{{(1+\frac{h}{a})}^n-1\}}{h}$

we can apply binomial expression of $(1+x{)}^n$ where |x| < 1, $\left|\frac{h}{a}\right|$ < 1

$(1+x{)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3+......$

When x infinitely small (approaching to zero) such that we can

ignore higher power of x, and have

$(1+x{)}^n=1+nx$ (approximately)

So, we can write

$=\underset{h\to 0}{lim}\frac{a^n\{1+\frac{nh}{a}-1\}}{h}$

$=\underset{h\to 0}{lim}\frac{a^n\{.\frac{n\text{/}h}{a}\}}{\text{/}h}$

$=\underset{h\to 0}{lim}\frac{a^n.n}{a}$

$=n{a}^{n-1}$

So,$\underset{x\to a}{lim}\frac{x^n-a^n}{x-a}=n{a}^{n-1}$

Option C is correct