Let f(1) = g(1) = k and their n^{th} derivati | vedclass

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(A) Given the limit: $\lim _{x \rightarrow 1} \frac{f(1) g(x) - f(1) - g(1) f(x) + g(1)}{g(x) - f(x)} = 4$. Since $f(1) = g(1) = k$,the expression bec

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$4$

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(A) Given the limit: $\lim _{x ightarrow 1} \frac{f(1) g(x) - f(1) - g(1) f(x) + g(1)}{g(x) - f(x)} = 4$. Since $f(1) = g(1) = k$,the expression becomes $\lim _{x ightarrow 1} \frac{k g(x) - k - k f(x) + k}{g(x) - f(x)} = \lim _{x ightarrow 1} \frac{k(g(x) - f(x))}{g(x) - f(x)}$. As $x ightarrow 1$,$g(x) - f(x) ightarrow g(1) - f(1) = k - k = 0$,which is an indeterminate form of type $\frac{0}{0}$. Applying $L'	ext{Hospital's Rule}$: $\lim _{x ightarrow 1} \frac{k g'(x) - k f'(x)}{g'(x) - f'(x)} = 4$. $\lim _{x ightarrow 1} \frac{k(g'(x) - f'(x))}{g'(x) - f'(x)} = 4$. Since $g'(1) 
eq f'(1)$ (as given that derivatives are not equal),we can cancel the term $(g'(x) - f'(x))$. Thus,$k = 4$.

Let $f(1) = g(1) = k$ and their $n^{th}$ derivatives $f^{(n)}(1), g^{(n)}(1)$ exist and are not equal for some $n$. If $\lim _{x \rightarrow 1} \frac{f(1) g(x) - f(1) - g(1) f(x) + g(1)}{g(x) - f(x)} = 4$,then the value of $k$ is:

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