Question

If $f\left(x\right)=\frac{2}{x-3}$, $g\left(x\right)=\frac{x-3}{x+4}$ and $h\left(x\right)=-\frac{2(2x+1)}{x^2+x-12}$, then $\underset{x\to 3}{lim}\left[f\right(x)+g(x)+h(x\left)\right]$ is

• A. $-2$
• B. $-1$
• C. $-\frac{2}{7}$
• D. $0$

Answer 1

Answer: (C)

$-\frac{2}{7}$

Given, $f\left(x\right)=\frac{2}{x-3}$, $g\left(x\right)=\frac{x-3}{x+4}$ and $h\left(x\right)=-\frac{2(2x+1)}{x^2+x-12}$
$f\left(x\right)+g\left(x\right)+h\left(x\right)=\frac{x^2-4x+17-4x-2}{x^2+x-12}=\frac{x^2-8x+15}{x^2+x-12}$
$\Rightarrow f\left(x\right)+g\left(x\right)+h\left(x\right)=\frac{(x-3)(x-5)}{(x-3)(x+4)}$
$\therefore \underset{x\to 3}{lim}\left[f\right(x)+g(x)+h(x\left)\right]$
$=\underset{x\to 3}{lim}\frac{(x-3)(x-5)}{(x-3)(x+4)}$
$=-\frac{2}{7}$