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 equal to:

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

Answer 1

Answer: (A)

$-\frac{2}{7}$

$\underset{x\to 3}{lim}\left[f\right(x)+g(x)+h(x\left)\right]$

$=\underset{x\to 3}{lim}[\frac{2}{x-3}+\frac{x-3}{x+4}+\frac{-2(2x+1)}{x^2+x-12}]$

$=\underset{x\to 3}{lim}\left[\frac{2(x+4)+(x-3{)}^2-2(2x+1)}{x^2+x-12}\right]$

$=\underset{x\to 3}{lim}\left[\frac{x^2-8x+15}{x^2+x-12}\right]=\underset{x\to 3}{lim}\left[\frac{(x-3)(x-5)}{(x+4)(x-3)}\right]$

$=\underset{x\to 3}{lim}\left[\frac{(x-5)}{(x+4)}\right]=-\frac{2}{7}$