Question

lim raiz de 2-t- raiz de 2/ t . Quando t tende à 0.

Answer 1

$\lim\limits_{t\to0}\dfrac{\sqrt{2-t}-\sqrt{2}}{t}$

Vamos tentar sumir com as raízes no numerador, multiplicando o numerador e o denominador da fração pelo "conjugado" do numerador, que é dado por $\sqrt{2-t}+\sqrt{2}$:

$\lim\limits_{t\to0}\dfrac{\sqrt{2-t}-\sqrt{2}}{t}=\lim\limits_{t\to0}\dfrac{\big(\sqrt{2-t}-\sqrt{2}\big)\big(\sqrt{2-t}+\sqrt{2}\big)}{t\big(\sqrt{2-t}+\sqrt{2})}$

Como $(a+b)(a-b)=(a-b)(a+b)=a^{2}-b^{2}$, temos

$\lim\limits_{t\to0}\dfrac{\sqrt{2-t}-\sqrt{2}}{t}=\lim\limits_{t\to0}\dfrac{(\sqrt{2-t})^{2}-(\sqrt{2})^{2}}{t\big(\sqrt{2-t}+\sqrt{2}\big)}\\\\\\\lim\limits_{t\to0}\dfrac{\sqrt{2-t}-\sqrt{2}}{t}=\lim\limits_{t\to0}\dfrac{(2-t)-2}{t\big(\sqrt{2-t}+\sqrt{2}\big)}\\\\\\\lim\limits_{t\to0}\dfrac{\sqrt{2-t}-\sqrt{2}}{t}=\lim\limits_{t\to0}\dfrac{-t}{t\big(\sqrt{2-t}+\sqrt{2}\big)}$

Como estudamos $t\neq0$ no limite, podemos cancelar $t$:

$\lim\limits_{t\to0}\dfrac{\sqrt{2-t}-\sqrt{2}}{t}=\lim\limits_{t\to0}\dfrac{-1}{\big(\sqrt{2-t}+\sqrt{2}\big)}$

Agora podemos substituir $t=0$ no limite:

$\lim\limits_{t\to0}\dfrac{\sqrt{2-t}-\sqrt{2}}{t}=\dfrac{-1}{\sqrt{2-0}+\sqrt{2}}\\\\\\\lim\limits_{t\to0}\dfrac{\sqrt{2-t}-\sqrt{2}}{t}=-\dfrac{1}{\sqrt{2}+\sqrt{2}}\\\\\\\boxed{\boxed{\lim\limits_{t\to0}\dfrac{\sqrt{2-t}-\sqrt{2}}{t}=-\dfrac{1}{2\sqrt{2}}}}$

Ou, se racionalizarmos,

$\boxed{\boxed{\lim\limits_{t\to0}\dfrac{\sqrt{2-t}-\sqrt{2}}{t}=\dfrac{\sqrt{2}}{4}}}$