Question

Evaluate: $\underset{x\to 3}{lim}\left[\frac{1}{x-3}+\frac{9x}{27-x^3}\right]$

Answer 1

$\begin{array}{c}{lim}_{x\to 3}[\frac{1}{x-3}+\frac{9x}{27-x^3}]\\ ={lim}_{x\to 3}\left[\frac{1}{x-3}-\frac{9x}{x^3-3^3}\right][\because a^3-b^3=(a-b\left)\right(a^2+ab+b^2\left)\right]\\ ={lim}_{x\to 3}\left[\frac{1}{x-3}-\frac{9x}{(x-3)(x^2+3x+9)}\right]\\ ={lim}_{x\to 3}\frac{x^2+3x+9-9x}{\left(x-3\right)\left(x^2+3x+9\right)}\\ ={lim}_{x\to 3}\frac{x^2-6x+9}{\left(x-3\right)\left(x^2+3x+9\right)}\\ ={lim}_{x\to 3}\frac{{\left(x-3\right)}^2}{\left(x-3\right)\left(x^2+3x+9\right)}\end{array}$

$=\underset{x\to 3}{lim}\frac{x-3}{x^3+3x+9}$

$=0$.