Respuesta :

[tex]f(x)=3x^2+3\hspace{5em}\displaystyle \lim_{h\to 0}~\cfrac{f(2+h)~~ - ~~f(2)}{h} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\cfrac{f(2+h)~~ - ~~f(2)}{h}\implies \cfrac{[3(2+h)^2+3]~~ - ~~[3(2)^2+3]}{h} \\\\\\ \cfrac{[3(4+4h+h^2)+3]~~ - ~~[3(4)+3]}{h}\implies \cfrac{[(12+12h+3h^2)+3]~~ - ~~[15]}{h} \\\\\\ \cfrac{[12h+3h^2+15]~ - ~~[15]}{h}\implies \cfrac{12h+3h^2}{h}\implies \cfrac{h(12+3h)}{h}\implies 12+3h \\\\[-0.35em] ~\dotfill\\\\ \displaystyle \lim_{h\to 0}~\cfrac{f(2+h)~~ - ~~f(2)}{h}\implies \lim_{h\to 0}~12 + 3h\implies 12+3(0)\implies \text{\LARGE 12}[/tex]