If $\underset{x\to a}{\lim }[f\left(x\right)+g\left(x\right)]=2$ and $\underset{x\to a}{\lim }[f\left(x\right)-g\left(x\right)]=1,$ then the value of $\underset{x\to a}{\lim }f\left(x\right)g\left(x\right)$ is

$\underset{x\to a}{\lim }[f\left(x\right)+g\left(x\right)]=2$ or $\underset{x\to a}{\lim }f\left(x\right)+\underset{x\to a}{\lim }g\left(x\right)=2\left(1\right)$

$\underset{x\to a}{\lim }[f\left(x\right)-g\left(x\right)]=1$ or $\underset{x\to a}{\lim }f\left(x\right)-\underset{x\to a}{\lim }g\left(x\right)=1\left(2\right)$

Adding (1) and (2), $2\underset{x\to a}{\lim }f\left(x\right)=3or\underset{x\to a}{\lim }f\left(x\right)=\frac{3}{2}$

Subtracting (2) from (I) 

$2\underset{x\to a}{\lim }g\left(x\right)=1or\underset{x\to a}{\lim g\left(x\right)}=\frac{1}{2} or \underset{x\to a}{\lim }f\left(x\right)g\left(x\right)=\underset{x\to a}{\lim }f\left(x\right)\underset{x\to a}{\lim }g\left(x\right)=\frac{3}{2}\times \frac{1}{2}=\frac{3}{4}$