Question

If $f(a) = 2$, $f^{'}(a) = 1$, $g(a) = - 3$, $g^{'}(a) = - 1$, then

$\lim_{x \rightarrow a}\frac{f(a)g(x) - f(x)g(a)}{x - a} =$

• A. 1
• B. 6
• C. – 5
• D. – 1

Answer 1

Answer: (A)

1

Answer 2

$\underset{x \rightarrow a}{\text{lim}}{}\frac{f(a)g(x) - f(x)g(a)}{x - a}$ $\left( \frac{0}{0}\text{form} \right)$

Using L-Hospital’s rule, $\lim _ { x \rightarrow a } \frac { f ( a ) g ^ { \prime } ( x ) - f ^ { \prime } ( x ) g ( a ) } { 1 - 0 }$

$= f(a) \times g^{'}(a) - f^{'}(a) \times g(a) = 2 \times ( - 1) - 1 \times ( - 3) = 1$.