Answer
$\dfrac{3}{2}$
Work Step by Step
Here, we have $f(x)=\sqrt{3x+1}; x=0$
and
$f(x+h)=\sqrt{3(x+h)+1}$
Let us multiply both numerator and denominator with $\sqrt{3(x+h)+1}+\sqrt{3x+1}$, and then we have
$\lim_{h\to0}\dfrac{f(x+h)f(x)}{h}=\lim\limits_{h\to0}\dfrac{3h}{h(\sqrt{3(x+h)+1}+\sqrt{3x+1})}=\lim\limits_{h\to0}\dfrac{3}{(\sqrt{3(x+h)+1}+\sqrt{3x+1})}$
Plug $x=0$ we get
$\lim\limits_{h\to0}\dfrac{3}{\sqrt{3(0+h)+1}+\sqrt{3\times0+1}}=\lim\limits_{h\to0}\dfrac{3}{\sqrt{3h+1}+\sqrt1}$
Thus, we have
$\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}=\dfrac{3}{\sqrt{3\times0+1}+1}=\dfrac{3}{2}$
