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
$\underset{x\to 3}{\lim }\left[f\left(x\right)+g\left(x\right)+h\left(x\right)\right]$ is = $-\frac{2}{y}$, find the value of $y$

Answer 1

Answer: 7

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