Suppose $f(x)$ is differentiable at $x = 1$ and $\lim_{h \rightarrow 0}{\frac{1}{h}f(1 + h) = 5}$, then $f^{'}(1)$ equals

Question

Dr. Abubacker · Accepted Answer

(b)
$f^{'}(1) = \lim_{h \rightarrow 0}\frac{f(1 + h) - f(1)}{h} = \lim_{h \rightarrow 0}\frac{f(1 + h)}{h} - \lim_{h \rightarrow 0}\frac{f(1)}{h}$
Given, $\lim_{h \rightarrow 0}\ \frac{f(1 + h)}{h} = 5$
So, $\lim_{h \rightarrow 0}\ \frac{f(1)}{h}$ must be finite as $f^{'}(1)$ exist and $\lim_{h \rightarrow 0}\ \frac{f(1)}{h}$ can be finite only, if $f(1) = 0$ and $\lim_{h \rightarrow 0}\ \frac{f(1)}{h} = 0$
So, $f^{'}(1) = \lim_{h \rightarrow 0}\ \frac{f(1 + h)}{h} = 5$

Q. Suppose $f(x)$ is differentiable at $x = 1$ and $\lim_{h \rightarrow 0}{\frac{1}{h}f(1 + h) = 5}$, then $f^{'}(1)$ equals

Explanation

Reviewed by Dr. Abubacker

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