Question

${lim}_{x\to 3}(\frac{1}{x-3}+\frac{9x}{27-x^2}).$

Answer 1

${lim}_{x\to 3}(\frac{1}{x-3}+\frac{9x}{27-x^3})=\frac{1}{x-3}+\frac{9x}{(3-x)}(9+x^2+3x)$

${lim}_{x\to 3}\frac{1}{(x-3)}\left(\frac{9+x^2+3x-9x}{9+x^2+3x}\right)$

$={lim}_{x\to 3}\frac{1}{(x-3)}\left[\frac{x^2-6x+9}{9+x^2+3x}\right]$

$={lim}_{x\to 3}\frac{(x-3{)}^2}{(x-3)(x^2+3x+9)}$

$={lim}_{x\to 3}\frac{(x-3)}{x^2+3x+9}=0$