## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\frac{3}{2}$$
$$f(x)=\sqrt{3x+1}\hspace{1cm} x=0$$ So $f(x+h)=\sqrt{3(x+h)+1}$ Therefore, $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{\sqrt{3(x+h)+1}-\sqrt{3x+1}}{h}$$ -Multiply both numerator and denominator with $\sqrt{3(x+h)+1}+\sqrt{3x+1}$, so that - Numerator: $(\sqrt{3(x+h)+1}-\sqrt{3x+1})(\sqrt{3(x+h)+1}+\sqrt{3x+1})=[3(x+h)+1]-[3x+1]=[3x+3h+1]-3x-1=3h$ - Denominator: $h(\sqrt{3(x+h)+1}+\sqrt{3x+1})$ That means $$\lim_{h\to0}\frac{f(x+h)f(x)}{h}=\lim_{h\to0}\frac{3h}{h(\sqrt{3(x+h)+1}+\sqrt{3x+1})}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{3}{(\sqrt{3(x+h)+1}+\sqrt{3x+1})}$$ - Substitute $x=0$ here: $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{3}{\sqrt{3(0+h)+1}+\sqrt{3\times0+1}}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{3}{\sqrt{3h+1}+\sqrt1}=\lim_{h\to0}\frac{3}{\sqrt{3h+1}+1}$$ Therefore, $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\frac{3}{\sqrt{3\times0+1}+1}=\frac{3}{\sqrt1+1}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\frac{3}{2}$$