Question

If $f(x) = \frac{2}{x - 3}, g(x) = \frac{x - 3}{x + 4},$ and $h(x) = - \frac{2(2x + 1)}{x^2 + x - 12}$, then
$\lim\limits_{x\to3} \left[f\left(x\right) + g\left(x\right) + h\left(x\right)\right]$ is = $-\frac{2}{y}$, find the value of $y$

Answer 1

We have
$f(x) + g(x) + h(x) = \frac{x^2 -4x+17 -4x -2}{ x^2 +x -12}$
$ = \frac{x^2 - 8x + 15}{x^2 + x -12} = \frac{(x-3)(x - 5)}{(x - 3)(x + 4)}$
$\therefore \lim\limits_{x\to3} \left[f\left(x\right) + g\left(x\right) + h\left(x\right)\right] = \lim\limits_{x\to 3} \frac{\left(x-3\right)\left(x-5\right)}{\left(x-3\right)\left(x+4\right)} $
$= -\frac{2}{7}$