lim √2x-√x+1 X→ 1 x-1.... Pergunta de ideia decarlanepereira

March 2022 1 39 Report

lim   √2x-√x+1

X→ 1   x-1

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Niiya

Verified answer

$\lim\limits_{x\rightarrow1}~\dfrac{\sqrt{2x}-\sqrt{x+1}}{x-1}=\dfrac{\sqrt{2\cdot1}-\sqrt{1+1}}{1-1}=\dfrac{0}{0}$

Quando temos limites chegando em indeterminações, podemos usar a regra de L'Hopital:

$\lim\limits_{x\rightarrow a}~\dfrac{f(x)}{g(x)}=\lim\limits_{x\rightarrow a}~\dfrac{f'(x)}{g'(x)}$

Onde f'(x) e g'(x) são, respectivamente, as derivadas de f(x) e g(x)
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$f(x)=\sqrt{2x}-\sqrt{x+1}$

Calculando a derivada de f(x) [Regra da cadeia]:

$f'(x)=\dfrac{1}{2\sqrt{2x}}\cdot1\cdot2x^{1-1}-\dfrac{1}{2\sqrt{x+1}}\cdot(1\cdot x^{1-1}+0)\\\\\\f'(x)=\dfrac{1}{\sqrt{2x}}-\dfrac{1}{2\sqrt{x+1}}$

Calculando a derivada de g(x):

$g'(x)=x-1\\g'(x)=1\cdot x^{1-1}-0\\g'(x)=1$

Logo, temos:

$\lim\limits_{x\rightarrow1}~\dfrac{\sqrt{2x}-\sqrt{x+1}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(\frac{1}{\sqrt{2x}}\right)-\left(\frac{1}{2\sqrt{x+1}}\right)}{1}\\\\\\\lim\limits_{x\rightarrow1}~\dfrac{\sqrt{2x}-\sqrt{x+1}}{x-1}=\dfrac{1}{\sqrt{2\cdot1}}-\dfrac{1}{2\sqrt{1+1}}\\\\\\\lim\limits_{x\rightarrow1}~\dfrac{\sqrt{2x}-\sqrt{x+1}}{x-1}=\dfrac{1}{\sqrt{2}}-\dfrac{1}{2\sqrt{2}}\\\\\\\lim\limits_{x\rightarrow1}~\dfrac{\sqrt{2x}-\sqrt{x+1}}{x-1}=\dfrac{2-1}{2\sqrt{2}}=\dfrac{1}{2\sqrt{2}}$

Se preferir, podemos racionalizar:

$\lim\limits_{x\rightarrow1}~\dfrac{\sqrt{2x}-\sqrt{x+1}}{x-1}=\dfrac{\sqrt{2}}{2\sqrt{2}\sqrt{2}}\\\\\\\lim\limits_{x\rightarrow1}~\dfrac{\sqrt{2x}-\sqrt{x+1}}{x-1}=\dfrac{\sqrt{2}}{2\cdot2}\\\\\\\boxed{\boxed{\lim\limits_{x\rightarrow1}~\dfrac{\sqrt{2x}-\sqrt{x+1}}{x-1}=\dfrac{\sqrt{2}}{4}}}$

1 votes Thanks 3

carlanepereira muito obrbigada

Niiya nada :)

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