Explanation
The limit definition of derivative is:
[tex]f^{\prime}(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}[/tex]
The given function is a constant function, which means that no matter what value is assigned to h, f(x+h) will always be equal to 3.
[tex]\begin{gathered} f^{\prime}(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h} \\ f^{\prime}(x)=\lim_{h\to0}\frac{3-3}{h} \\ f^{\prime}(x)=\lim_{h\to0}\frac{0}{h} \end{gathered}[/tex]
Now, we evaluate the limit of the numerator and the denominator.
[tex]\begin{gathered} f^{\prime}(x)=\frac{\lim_{h\to0}0}{\lim_{h\to0}h} \\ f^{\prime}(x)=\frac{0}{0} \end{gathered}[/tex]
Since we have an indeterminate form, we apply the L'Hospital's Rule which states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.
[tex]\begin{gathered} f^{\prime}(x)=\lim_{h\to0}\frac{0}{h} \\ f^{\prime}(x)=\lim_{h\to0}\frac{\frac{d}{dh}0}{\frac{d}{dh}h} \\ f^{\prime}(x)=\lim_{h\to0}\frac{0}{1} \\ f^{\prime}(x)=\lim_{h\to0}0 \end{gathered}[/tex]
Finally, the limit as h approaches 0 is 0.
[tex]\begin{gathered} f^{\prime}(x)=\lim_{h\to0}0 \\ f^{\prime}(x)=0 \end{gathered}[/tex]Answer[tex]f^{\prime}(x)=0[/tex]
