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 $\underset{x\to a}{lim}f\left(x\right)g\left(x\right)$

• A. need not exist
• B. exists and is $\frac{3}{4}$
• C. exists and is $-\frac{3}{4}$
• D. exists and is $\frac{4}{3}$

Answer 1

The correct option is B exists and is $\frac{3}{4}$

Since the sum and subtraction of two limits exists together, so their distinct limit will also exists.

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

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

Again

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

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

Solving (1) and (2) we get,

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

Since limits of $f\left(x\right)$ and $g\left(x\right)$ exists then their product also exists.

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

So, option (B) is true.