Question

If $x=3$, find the limit of $\frac{x^3-5x^2+7x-3}{x^3-x^2-5x-3}$.

Answer 1

To find the limit of $\frac{x^3-5x^2+7x-3}{x^3-x^2-5x-3}$ when $x=3$

i.e. $\underset{x\to 3}{lim}\frac{x^3-5x^2+7x-3}{x^3-x^2-5x-3}$

$=\underset{x\to 3}{lim}\frac{(x-3)(x^2-2x+1)}{(x-3)(x^2+2x+1)}$

$=\underset{x\to 3}{lim}\frac{(x^2-2x+1)}{(x^2+2x+1)}$

$=\frac{(3^2-2(3)+1)}{(3^2+2(3)+1)}$

$=\frac{4}{16}$

$=\frac{1}{4}$