Question

lim (x+h)^3 - x^3/ h quando x tende a 0

Answer 1

(x³+x²h+x²h+xh²+h²x+h³)-x³/h

h(2hx²+2xh²+h³)/h

=2x²+2xh+h³

x--> 0

= 2x0²+2x0xh²+h³
=h³

Answer 2

Vamos fatora  (x +h)³ pelo método binomial de newton.
                                                      n
formula: $(a +b)^n =$ → ∑      $a^n^-^k*b^k*c (n,k)$
                                                      k = 0

(x +h)³ = x³⁻⁰*h⁰*C3,0 + x³⁻¹*h¹*C3,1 + x³⁻²*h²*C3,2 + x³⁻³*h³*C3,3

(x +h)³ = x³C3,0 + x²hC3,1 + xh²C3,2 + h³C3,3

C3,0 = $\frac{3!}{0!3!} = 1$

C3,1 = $\frac{3!}{1!(3-1)!} = \frac{3*2!}{1*2!} = 3$

C3,2 = $\frac{3!}{2!(3-2)!} = \frac{3*2!}{2!1"} = 3$

C3,3 = $\frac{3!}{0!3!} = 1$


(x + h)³ = 1x³ + 3x²h + 3xh² + 1h³

Vamos para o limite!

$\\ \lim_{x \to 0} x^3 + 3x^2h +3xh^2 +h^3 - \frac{x^3}{h} = \\ \\ \lim_{x \to 0} \frac{h(x^3 + 3x^2h +3xh^2 + h^3) - x^3}{h} = \\ \\ \lim_{x \to 0} \frac{x^3h +3x^2h^2+3xh^3+h^4-x^3}{h} = \\ \\ \lim_{x \to 0} \frac{0^3*h+3*0^2*h +3*0*h^3+h^4-0^3}{h} = \\ \\ \lim_{x \to 0} \frac{h^4}{h} = \\ \\ \lim_{x \to 0} h^4-^1 = h^3$