Question

Assertion :If $\underset{x\to 0}{lim}f\left(x\right)$ and $\underset{x\to 0}{lim}g\left(x\right)$ exists finitely, then $\underset{x\to 0}{lim}f\left(x\right)\cdot g\left(x\right)$ exists finitely. Reason: If $\underset{x\to 0}{lim}f\left(x\right)\cdot g\left(x\right)$ exists finitely then $\underset{x\to 0}{lim}f\left(x\right)\cdot g\left(x\right)=\underset{x\to 0}{lim}f\left(x\right)\cdot \underset{x\to 0}{lim}g\left(x\right)$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect and Reason is correct

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

From the property of limits, the limit of a product is the product of the limits provided that the two limits on
the right side are defined.
$\because {lim}_{x\to 0}f\left(x\right)=L$ and ${lim}_{x\to 0}g\left(x\right)=M$
Using above property, ${lim}_{x\to 0}f\left(x\right)g\left(x\right)={lim}_{x\to 0}f\left(x\right).{lim}_{x\to 0}g\left(x\right)=L.M$
Thus, assertion is correct.
But the reason is not correct always. Let us consider following situation
$f\left(x\right)=x$ and $g\left(x\right)=\frac{1}{x}$
$\Rightarrow {lim}_{x\to 0}f\left(x\right).g\left(x\right)={lim}_{x\to 0}x.\frac{1}{x}=1$
But ${lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}x=0$ and ${lim}_{x\to 0}g\left(x\right)={lim}_{x\to 0}\frac{1}{x}$ is not defined at $x\to 0$
Thus reason is incorrect.