Let $f(a) = g(a) = k$ and their $n$th derivatives $f^{n}(a),\ g^{n}(a)$ exist and are not equal for some$n$. Further if $\lim_{x \rightarrow a}\frac{f(a)g(x) - f(a) - g(a)f(x) + g(a)}{g(x) - f(x)} = 4$, then the value of $k$ is equal to

Question

Dr. Abubacker · Accepted Answer

(a)
Given, $\lim_{x \rightarrow a}\frac{f(a)g(x) - f(a) - g(a)f(x) + g(a)}{g(x) - f(x)} = 4$
Applying L’ Hospital’s rule,
$\Rightarrow \ \ \ \ \ \ \lim_{x \rightarrow a}\frac{f(a)g'(x) - g(a)f'(x)}{g'(x) - f'(x)} = 4$
$\Rightarrow \ \ \ \ \ \ \ \ \lim_{x \rightarrow a}\frac{kg^{'(x)} - kf'(x)}{g'(x) - f'(x)} = 4$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k = 4$

Q. Let $f(a) = g(a) = k$ and their $n$th derivatives $f^{n}(a),\ g^{n}(a)$ exist and are not equal for some$n$. Further if $\lim_{x \rightarrow a}\frac{f(a)g(x) - f(a) - g(a)f(x) + g(a)}{g(x) - f(x)} = 4$, then the value of $k$ is equal to

Explanation

Reviewed by Dr. Abubacker

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