Question

calcule o limite de f(x)= (2+X)²- 4 / X quando x=0 é

Answer 1

$\lim\limits_{x\rightarrow0}\dfrac{(2+x)^{2}-4}{x}=\lim\limits_{x\rightarrow0}\dfrac{(2+x)^{2}-2^{2}}{x}$

A diferença de quadrados pode ser escrita da seguinte forma:

$a^{2}-b^{2}=(a+b)(a-b)~~(produto~da~soma~pela~dif.~de~dois~termos)$

Logo:

$\lim\limits_{x\rightarrow0}\dfrac{(2+x)^{2}-4}{x}=\lim\limits_{x\rightarrow0}\dfrac{(2+x+2)(2+x-2)}{x}\\\\\\\lim\limits_{x\rightarrow0}\dfrac{(2+x)^{2}-4}{x}=\lim\limits_{x\rightarrow0}\dfrac{(4+x)(x)}{x}$

Cortando x:

$\lim\limits_{x\rightarrow0}\dfrac{(2+x)^{2}-4}{x}=\lim\limits_{x\rightarrow0}(4+x)\\\\\\\lim\limits_{x\rightarrow0}\dfrac{(2+x)^{2}-4}{x}=4+0\\\\\\\boxed{\boxed{\lim\limits_{x\rightarrow0}\dfrac{(2+x)^{2}-4}{x}=4}}$

Answer 2

$\lim\limits_{x\to0}\dfrac{(2+x)^2-4}{x}=\lim\limits_{x\to0}\dfrac{x^2+4x}{x}=\lim\limits_{x\to0}\dfrac{x(x+4)}{x}=\\\\\\=\lim\limits_{x\to0}x+4=0+4=\bold{4}$