Let f : R $\to$ R satisfy the equation f(x + y) = f(x) . f(y) for all x, y $\in$R and f(x) $\ne$ 0 for any x$\in$R. If the function f is differentiable at x = 0 and f'(0) = 3, then
$\mathop {\lim }\limits_{h \to 0} {1 \over h}(f(h) - 1)$ is equal to ____________.
Answer (integer)
3
Solution
<p>Given, $f(x + y) = f(x)\,.\,f(y)\,\forall x,y \in R$</p>
<p>$\therefore$ $f(x) = {a^x} \Rightarrow f'(x) = {a^x}\,.\,\log (a)$</p>
<p>Now, $f'(0) = \log (a) \Rightarrow 3 = \log (a) \Rightarrow a = {e^3}$</p>
<p>$\therefore$ $f(x) = {({e^3})^x} = {e^{3x}}$</p>
<p>$\therefore$ $f(h) = {e^{3h}}$</p>
<p>Now, $$\mathop {\lim }\limits_{h \to 0} \left( {{{f(h) - 1} \over h}} \right) = \mathop {\lim }\limits_{h \to 0} \left( {{{{e^{3h}} - 1} \over h}} \right)$$</p>
<p>$$ = \mathop {\lim }\limits_{h \to 0} \left( {{{{e^{3h}} - 1} \over {3h}} \times 3} \right) = 3 \times 1 = 3$$</p>
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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