Question

Lim quando x tende a dois 1 sobre x menos 1 sobre 2 dividido por x ao quadrado menos 4

Answer 1

Usaremos o produto da soma pela diferença de dois termos (um produto notável):

$\boxed{\boxed{a^{2}-b^{2}=(a+b)\cdot(a-b)}}$
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$\lim\limits_{x\to2}\dfrac{\frac{1}{x}-\frac{1}{2}}{x^{2}-4}$

Vamos simplificar o numerador:

$\lim\limits_{x\to2}\dfrac{\frac{1}{x}-\frac{1}{2}}{x^{2}-4}=\lim\limits_{x\to2}\dfrac{\frac{2}{2x}-\frac{x}{2x}}{x^{2}-4}\\\\\\\lim\limits_{x\to2}\dfrac{\frac{1}{x}-\frac{1}{2}}{x^{2}-4}=\lim\limits_{x\to2}\dfrac{\big(\frac{2-x}{2x}\big)}{x^{2}-4}\\\\\\\lim\limits_{x\to2}\dfrac{\frac{1}{x}-\frac{1}{2}}{x^{2}-4}=\lim\limits_{x\to2}\dfrac{2-x}{2x}\cdot\dfrac{1}{x^{2}-4}\\\\\\\lim\limits_{x\to2}\dfrac{\frac{1}{x}-\frac{1}{2}}{x^{2}-4}=\lim\limits_{x\to2}\dfrac{2-x}{2x(x^{2}-4)}$

Agora, vamos fatorar $x^{2}-4$, usando o produto notável apresentado:

$\boxed{\boxed{x^{2}-4=x^{2}-2^{2}=(x+2)(x-2)}}$

Substituindo no limite:

$\lim\limits_{x\to2}\dfrac{\frac{1}{x}-\frac{1}{2}}{x^{2}-4}=\lim\limits_{x\to2}\dfrac{2-x}{2x(x^{2}-4)}\\\\\\\lim\limits_{x\to2}\dfrac{\frac{1}{x}-\frac{1}{2}}{x^{2}-4}=\lim\limits_{x\to2}\dfrac{2-x}{2x(x+2)(x-2)}\\\\\\\lim\limits_{x\to2}\dfrac{\frac{1}{x}-\frac{1}{2}}{x^{2}-4}=\lim\limits_{x\to2}\dfrac{-(x-2)}{2x(x+2)(x-2)}$

Cancelando $(x-2)$:

$\lim\limits_{x\to2}\dfrac{\frac{1}{x}-\frac{1}{2}}{x^{2}-4}=\lim\limits_{x\to2}\dfrac{-1}{2x(x+2)}$

Agora, podemos substituir $x=2$ na expressão, ficando com

$\lim\limits_{x\to2}\dfrac{\frac{1}{x}-\frac{1}{2}}{x^{2}-4}=\dfrac{-1}{2\cdot2\cdot(2+2)}\\\\\\\lim\limits_{x\to2}\dfrac{\frac{1}{x}-\frac{1}{2}}{x^{2}-4}=\dfrac{-1}{4\cdot4}\\\\\\\boxed{\boxed{\lim\limits_{x\to2}\dfrac{\frac{1}{x}-\frac{1}{2}}{x^{2}-4}=-\dfrac{1}{16}}}$