Question

usando a definição, calcule a derivada f (x) = 1/x com a=2

Answer 1

Pela definição,

$f'(2)=\lim\limits_{x\to2}\dfrac{f(x)-f(2)}{x-2}$

se o limite existir.
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Mas, note que

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

$\lim\limits_{x\to2}\dfrac{f(x)-f(2)}{x-2}=\lim\limits_{x\to2}\,\,\dfrac{-(x-2)}{2x\cdot(x-2)}$

Como $x\to2~~\Rightarrow~~x\neq2~~\Rightarrow~x-2\neq0$, então podemos cancelar $x-2$, ficando com

$\lim\limits_{x\to2}\dfrac{f(x)-f(2)}{x-2}=\lim\limits_{x\to2}-\dfrac{1}{2x}$

Agora podemos substituir $x=2$ normalmente:

$\lim\limits_{x\to2}\dfrac{f(x)-f(2)}{x-2}=-\dfrac{1}{2\cdot2}\\\\\\\boxed{\boxed{\lim\limits_{x\to2}\dfrac{f(x)-f(2)}{x-2}=-\dfrac{1}{4}}}$

Portanto, o limite existe, e

$\boxed{\boxed{f'(2)=-\dfrac{1}{4}}}$