Answer
$$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=3$$
Work Step by Step
$$f(x)=3x-4\hspace{1cm} x=2$$
So $f(x+h)=3(x+h)-4=3x+3h-4$
Therefore, $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{(3x+3h-4)-(3x-4)}{h}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{3x+3h-4-3x+4}{h}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{3h}{h}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}3$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=3$$
