Question

If $f\left(a\right)=2,f^{\prime }\left(a\right)=1,g\left(a\right)=3,g^{\prime }\left(a\right)=-1$ then $\underset{x\to a}{lim}\frac{f\left(a\right)g\left(x\right)-f\left(x\right)g\left(a\right)}{x-a}$ is equal to

• A. 6
• B. 1
• C. -1
• D. -5

Answer 1

The correct option is D -5

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

Using L'Hospital rule, we get the limit to be equal to $\underset{x\to a}{lim}\frac{f\left(a\right)g^{\prime }\left(x\right)-f^{\prime }\left(x\right)g\left(a\right)}{1}=f\left(a\right)g^{\prime }\left(a\right)-f^{\prime }\left(a\right)g\left(a\right)$

$=2\times (-1)-(1\times 3)=-5$