Question

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

Answer 1

Given ${lim}_{x\to a}\frac{x^3-a^3}{x-a}={lim}_{x\to 1}\frac{x^4-1}{x-1}.....\left(i\right)$

$LHS=If{lim}_{x\to a}\frac{x^3-a^3}{x-a}3(a{)}^{3-1}$

$=3a^2......\left(ii\right)$

$RHS={lim}_{x\to a}\frac{x^4-1}{x-1}$

$={lim}_{x\to a}\frac{x^4-1}{x-1}=4(1{)}^{4-1}$

=4 ......(iii)

Subsitituting (ii)and (iii)in (i),

$3a^2=4$

$a^2=\frac{4}{3}$

$a=\pm \frac{2}{\sqrt{3}}$