Matemática, perguntado por delgadonetoo, 1 ano atrás

lim 2-raiz de x-3/x^2-49 ten a7

Soluções para a tarefa

Respondido por Niiya
8
\lim\limits_{x\rightarrow7}~\dfrac{2-\sqrt{x-3}}{x^{2}-49}=\dfrac{2-\sqrt{7-3}}{7^{2}-49}=\dfrac{2-2}{49-49}=\dfrac{0}{0}

Se chegarmos a 0/0 ou a ∞/∞ na tentativa de resolução de um limite, podemos usar a regra de L'Hopital:

\boxed{\boxed{\lim\limits_{x\rightarrow a}~\dfrac{g(x)}{h(x)}=\lim\limits_{x\rightarrow a}~\dfrac{g'(x)}{h'(x)}}}
___________________

g(x)=2-\sqrt{x-3}

Derivando g(x) [a raiz será derivada pela regra da cadeia]

\dfrac{d}{dx}g(x)=\dfrac{d}{dx}2-\dfrac{d}{dx}\sqrt{x-3}\\\\\\g'(x)=0-\dfrac{1}{2\sqrt{x-3}}\cdot\dfrac{d}{dx}(x-3)\\\\\\\boxed{g'(x)=-\dfrac{1}{2\sqrt{x-3}}}

h(x)=x^{2}-49

Derivando h(x):

h'(x)=2\cdot x^{1-1}-0=2x

Logo:

\lim\limits_{x\rightarrow7}~\dfrac{2-\sqrt{x-3}}{x^{2}-49}=\lim\limits_{x\rightarrow7}~\dfrac{(-\frac{1}{2\sqrt{x+3}})}{2x}\\\\\\\lim\limits_{x\rightarrow7}~\dfrac{2-\sqrt{x-3}}{x^{2}-49}=\dfrac{(-\frac{1}{2\sqrt{7-3}})}{2\cdot7}\\\\\\\lim\limits_{x\rightarrow7}~\dfrac{2-\sqrt{x-3}}{x^{2}-49}=\dfrac{(-\frac{1}{2\sqrt{4}})}{14}\\\\\\\lim\limits_{x\rightarrow7}~\dfrac{2-\sqrt{x-3}}{x^{2}-49}=\dfrac{(-\frac{1}{2\cdot2})}{14}=-\dfrac{1}{4\cdot14}=-\dfrac{1}{56}
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