Question

Let $f:R\to R$ satisfy the equation $f(x+y)=f\left(x\right).f\left(y\right)$ for all $x,y\in R$ and $f\left(x\right)\ne 0$ for any $x\in R.$ If the function $f$ is differentiable at $x=0$ and $f\text{'}\left(0\right)=3,$then $\underset{h\to 0}{\lim }\frac{1}{h}\left[f\left(h\right)-1\right]$ is equal to:

Answer 1

Determine the value of $\underset{h\to 0}{\lim }\frac{1}{h}\left[f\left(h\right)-1\right]$

Given that, $f(x+y)=f\left(x\right).f\left(y\right)$, $f\text{'}\left(0\right)=3$

It is also given that: $f\left(x\right)\ne 0$

Putting $x=0,y=0$ in the equation:

$$\begin{gathered} f(0+0)=f\left(0\right).f\left(0\right) \\ \Rightarrow f\left(0\right)=1 \end{gathered}$$

Now,

$$\begin{gathered} \underset{h\to 0}{\lim }\frac{1}{h}\left[f\left(h\right)-1\right] \\ =\underset{h\to 0}{\lim }\frac{\left[f(h+0)-f\left(0\right)\right]}{h} \end{gathered}$$

$$\begin{gathered} \text{According to the first principle of derivative:} \\ f\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{\left[f(h+0)-f\left(0\right)\right]}{h} \\ \Rightarrow 3=\underset{h\to 0}{\lim }\frac{\left[f(h+0)-f\left(0\right)\right]}{h}...\left[Given\right] \\ \Rightarrow 3=\underset{h\to 0}{\lim }\frac{\left[f\left(h\right)-1\right]}{h} \\ \Rightarrow \underset{h\to 0}{\lim }\frac{\left[f\left(h\right)-1\right]}{h}=3 \end{gathered}$$

Hence, the value of $\underset{h\to 0}{\lim }\frac{1}{h}\left[f\left(h\right)-1\right]$ is $3$