Question

Evaluate:
$\begin{array}{c}\left(i\right)\underset{x\to 1}{lim}\frac{3x^3+5x-8}{2x^3+x-3}\\ \left(ii\right)\underset{x\to \frac{1}{2}}{lim}\frac{8x^3-1}{4x^3-x}\\ \left(iii\right)\underset{x\to 3}{lim}[\frac{1}{x-3}-\frac{9x}{x^3-27}]\end{array}$

Answer 1

(i) $\underset{x\to 1}{lim}\frac{{3x}^3+5x-8}{2x^3+x-3}$

$=\underset{x\to 1}{lim}\frac{9x^2+5}{6x^2+x}=\frac{14}{7}=2$ [by L-Hospital's Rule]

(ii) $\underset{x\to 1/2}{lim}\frac{{8x}^3-1}{4x^3-x}$

$=\underset{x\to 1/2}{lim}\frac{{24x}^2}{{12x}^2-1}$ [L-Hospital's Rule]

$=\frac{24-\frac{1}{4}}{12-\frac{1}{4}-1}=\frac{6}{2}=3$

(iii) $\underset{x\to 3}{lim}[\frac{1}{x-3}-\frac{9x}{x^3-27}]$

$=\underset{x\to 3}{lim}[\frac{x^2+3x+9}{(x-3)(x^2+3x+9)}-\frac{9x}{x^3-27}]$

$=\underset{x\to 3}{lim}\left[\frac{x^2+9-6x}{x^3-27}\right]$

$=\underset{x\to 3}{lim}\left[\frac{{(x-3)}^2}{(x-3)(x^2+3x+9)}\right]$

$=\underset{x\to 3}{lim}\left[\frac{{(x-3)}^{}}{(x^2+3x+9)}\right]=0$