Question

If $\underset{x\to a}{lim}\frac{x^3-a^3}{x-a}=\underset{x\to 1}{lim}\frac{x^4-1}{x-1}$, find all possible values of a.

Answer 1

$\underset{x\to a}{lim}\frac{x^3-a^3}{x-a}=\frac{0}{0}$ which is the indeterminate form

Therefore, L-hospitals rule is applied. i.e.,we take the derivative of numerator and denominator

$\underset{x\to a}{lim}\frac{\frac{d}{dx}(x^3-a^3)}{\frac{d}{dx}(x-a)}=\underset{x\to a}{lim}3x^2=3a^2$

$\underset{x\to 1}{lim}\frac{x^4-1}{x-1}=\frac{0}{0}$ which is the indeterminate form

Therefore, L-hospitals rule is applied. i.e.,we take the derivative of numerator and denominator

$\underset{x\to 1}{lim}\frac{\frac{d}{dx}(x^4-1)}{\frac{d}{dx}(x-1)}=\underset{x\to 1}{lim}4x^3=4$

Given that $\underset{x\to a}{lim}\frac{x^3-a^3}{x-a}=\underset{x\to 1}{lim}\frac{x^4-1}{x-1}$

Therefore, $3a^2=4$

$⟹a^2=\frac{4}{3}$

$⟹a=\sqrt{\frac{4}{3}}$