If $f(a) = 2,\ \ f^{'}(a) = 1,\ g(a) = 3,\ g^{'}(a) = - 1$, then $\lim_{x \rightarrow a}\frac{f(a)g(x) - f(x)g(a)}{x - a}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(d)
$\lim_{x \rightarrow a}\frac{f(a)g(x) - f(x)g(a)}{x - a}$ $\left\lbrack \frac{0}{0}\ from \right\rbrack$
$= \lim_{x \rightarrow a}\frac{f(a)g'(x) - f'(x)g(a)}{1 - 0}$ [by L’ Hospital’s rule]
$= f(a)g'(a) - f'(a)g(a)$
$= 2( - 1) - 1(3) = - 2 - 3 = - 5$

Q. If $f(a) = 2,\ \ f^{'}(a) = 1,\ g(a) = 3,\ g^{'}(a) = - 1$, then $\lim_{x \rightarrow a}\frac{f(a)g(x) - f(x)g(a)}{x - a}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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