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 $\underset{x\to 3}{lim 7} \left\{f\right(x)+g(x)+h(x\left)\right\}$ is

Solution:

We have,

$\begin{array}{r}f\left(x\right)+g\left(x\right)+h\left(x\right)=\frac{x^2-4x+17-4x-2}{x^2+x-12}=\frac{(x-3)(x-5)}{(x-3)(x+4)}\\ \therefore \underset{x\to 3}{lim} \left[f\right(x)+g(x)+h(x\left)\right]=\underset{x\to 3}{lim} \frac{(x-3)(x-5)}{(x-3)(x+4)}=-\frac{2}{7}\end{array}$