Question

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

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{3}{4}$
• D. $\frac{4}{5}$

Answer 1

The correct option is C $\frac{3}{4}$

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

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

Adding (1) and (2),
$2\underset{x\to a}{lim}f\left(x\right)=3$
$\underset{x\to a}{lim}f\left(x\right)=\frac{3}{2}$
Subtracting (2) from (1),
$2\underset{x\to a}{lim}g\left(x\right)=1$
$\underset{x\to a}{lim}g\left(x\right)=\frac{1}{2}$
Now, $\underset{x\to a}{lim}f\left(x\right)g\left(x\right)$
$=\underset{x\to a}{lim}f\left(x\right)\underset{x\to a}{lim}g\left(x\right)$
$=\frac{3}{2}\times \frac{1}{2}=\frac{3}{4}$