Question

Lim ( raiz de 1+x)-( raiz de 1-x)/x quando x tende a 0

Answer 1

$\boxed{ \lim_{x \to 0} \frac{ \sqrt{1+x}- \sqrt{1-x} }{x} = \frac{0}{0} }$


imagina assim
$A = \sqrt{1+x} \\\\\ B= \sqrt{x-1}$

vc teria
$\frac{A-B}{x}$

multiplica o numerador e o denominador  pelo conjugado
conjugado de A-B = A+B
então fica

$\lim_{x \to 0} \frac{(A-B)*(A+B)}{x*(A+B)}$

no numerador vc tem uma diferença dos quadrados
$(A-B)*(A+B)= A^2-B^2$

ficando com
$\lim_{x \to 0} \frac{A^2 -B^2}{x*(A+B)} \\\\ = \lim_{x \to 0} \frac{( \sqrt{1+x} )^2- (\sqrt{1-x})^2 }{x*( \sqrt{1+x}+ \sqrt{1-x}) } \\\\ \lim_{x \to 0} \frac{1+x -(1-x)}{x*( \sqrt{1+x}+ \sqrt{1-x}) } \\\\ = \lim_{x \to 0} \frac{2\not x}{\not x*( \sqrt{1+x}+ \sqrt{1-x}) } \\\\ = \lim_{x \to 0} \frac{2}{ \sqrt{1+x}+ \sqrt{1-x} }= \frac{2}{ \sqrt{1+0}+ \sqrt{1-0} }= \frac{2}{1+1}= 1$

Answer 2

Resolvendo de outro jeito:

$\lim\limits_{x\rightarrow 0}~\dfrac{\sqrt{1+x}-\sqrt{1-x}}{x}$

Se substituirmos x por 0, chegamos numa indeterminação 0/0, então podemos usar a Regra de L'Hopital:

$\boxed{\boxed{\lim\limits_{x\rightarrow a}~\dfrac{f(x)}{g(x)}=\lim\limits_{x\rightarrow a}~\dfrac{f'(x)}{g'(x)}}}$
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Vamos chamar o numerador de f(x) e o denominador de g(x)

Achando a derivada de f(x):

$f(x)=\sqrt{1+x}-\sqrt{1-x}$

Acharei as derivadas dessas raízes pela regra da cadeia:

$f'(x)=\dfrac{1}{2\sqrt{1+x}}\cdot\dfrac{d}{dx}(1+x)-\dfrac{1}{2\sqrt{1-x}}\cdot\dfrac{d}{dx}(1-x)\\\\\\f'(x)=\dfrac{1}{2\sqrt{1+x}}\cdot1-\dfrac{1}{2\sqrt{1-x}}\cdot(-1)\\\\\\\boxed{\boxed{f'(x)=\dfrac{1}{2\sqrt{1+x}}+\dfrac{1}{2\sqrt{1-x}}}}$

Achando a derivada de g(x):

$g(x)=x\\g'(x)=1$
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$\lim\limits_{x\rightarrow 0}~\dfrac{\sqrt{1+x}-\sqrt{1-x}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\frac{1}{2\sqrt{1+x}}+\frac{1}{2\sqrt{1-x}}}{1}\\\\\\\lim\limits_{x\rightarrow 0}~\dfrac{\sqrt{1+x}-\sqrt{1-x}}{x}=\dfrac{1}{2\sqrt{1+0}}+\dfrac{1}{2\sqrt{1-0}}\\\\\\\lim\limits_{x\rightarrow 0}~\dfrac{\sqrt{1+x}-\sqrt{1-x}}{x}=\dfrac{1}{2}+\dfrac{1}{2}\\\\\\\boxed{\boxed{\lim\limits_{x\rightarrow 0}~\dfrac{\sqrt{1+x}-\sqrt{1-x}}{x}=1}}$