Question

Me ajudem com esses limites, isso não é o meu forte.​

Answer 1

Resposta:

Explicação passo-a-passo:

m) Multiplicar o numerador e denominador pelo conjugado do numerador

$\lim_{x\to 0}\dfrac{(\sqrt{1-2x-x^{2}}-1)(\sqrt{1-2x-x^{2}}+1)}{x(\sqrt{1-2x-x^{2}}+1)}=\\\\\\=\lim_{x\to 0}\dfrac{1-2x-x^{2}-1}{x(\sqrt{1-2x-x^{2}}+1)}=\\\\\\ =\lim_{x\to 0}\dfrac{-2x-x^{2}}{x(\sqrt{1-2x-x^{2}}+1)}=\\\\\\ =\lim_{x\to 0}\dfrac{x(-2-x)}{x(\sqrt{1-2x-x^{2}}+1)}=\\\\\\=\lim_{x\to 0}\dfrac{-2-x}{\sqrt{1-2x-x^{2}}+1}=\\\\\\=\dfrac{-2-0}{\sqrt{1-2.0-0^{2}}+1}=\dfrac{-2}{\sqrt{1}+1}=\dfrac{-2}{1+1}=\dfrac{-2}{2}=-1$

n) Multiplicar o numerador e denominador pelo conjugado do numerador

$\lim_{x\to 0}\dfrac{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}{x(\sqrt{1+x}+\sqrt{1-x})}=\\\\\\= \lim_{x\to 0}\dfrac{1+x-(1-x)}{x(\sqrt{1+x}+\sqrt{1-x})}=\\\\\\= \lim_{x\to 0}\dfrac{1+x-1+x}{x(\sqrt{1+x}+\sqrt{1-x})}=\\\\\\= \lim_{x\to 0}\dfrac{x+x}{x(\sqrt{1+x}+\sqrt{1-x})}=\\\\\\= \lim_{x\to 0}\dfrac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}=\\\\\\= \lim_{x\to 0}\dfrac{2}{\sqrt{1+x}+\sqrt{1-x}}=\\\\\\=\dfrac{2}{\sqrt{1+0}+\sqrt{1-0}}=\dfrac{2}{\sqrt{1}+\sqrt{1}}=\dfrac{2}{1+1}=\dfrac{2}{2}=1$

o) Multiplicar o numerador e denominador pelo conjugado do numerador

$\lim_{x\to 1}\dfrac{(\sqrt{2x}-\sqrt{x+1})(\sqrt{2x}+\sqrt{x+1})}{(x-1)(\sqrt{2x}+\sqrt{x+1})}=\\\\\\=\lim_{x\to 1}\dfrac{2x-(x+1)}{(x-1)(\sqrt{2x}+\sqrt{x+1})}=\\\\\\=\lim_{x\to 1}\dfrac{2x-x-1}{(x-1)(\sqrt{2x}+\sqrt{x+1})}=\\\\\\=\lim_{x\to 1}\dfrac{(x-1)}{(x-1)(\sqrt{2x}+\sqrt{x+1})}=\\\\\\=\lim_{x\to 1}\dfrac{1}{\sqrt{2x}+\sqrt{x+1}}=\\\\\\=\dfrac{1}{\sqrt{2.1}+\sqrt{1+1}}=\dfrac{1}{\sqrt{2}+\sqrt{2}}=\\\\=\dfrac{1}{2\sqrt{2}}=\dfrac{\sqrt{2}}{2\sqrt{2}\sqrt{2}}=\dfrac{\sqrt{2}}{2.2}=$

$=\dfrac{\sqrt{2}}{4}$

p) Multiplicar o numerador e denominador pelo conjugado do denominador

$\lim_{x\to 2} \dfrac{(x^{2}-4)(\sqrt{x+2}+\sqrt{3x-2})}{(\sqrt{x+2}-\sqrt{3x-2})(\sqrt{x+2}+\sqrt{3x-2})}=\\\\\\ =\lim_{x\to 2} \dfrac{(x^{2}-4)(\sqrt{x+2}+\sqrt{3x-2})}{x+2-(3x-2)}=\\\\\\ =\lim_{x\to 2} \dfrac{(x^{2}-4)(\sqrt{x+2}+\sqrt{3x-2})}{x+2-3x+2}=\\\\\\ =\lim_{x\to 2} \dfrac{(x-2)(x+2)(\sqrt{x+2}+\sqrt{3x-2})}{-2x+4}=\\\\\\ =\lim_{x\to 2} \dfrac{(x-2)(x+2)(\sqrt{x+2}+\sqrt{3x-2})}{-2(x-2)}=\\\\\\ =\lim_{x\to 2} \dfrac{(x+2)(\sqrt{x+2}+\sqrt{3x-2})}{-2}=$

$=-\dfrac{(2+2)(\sqrt{2+2}+\sqrt{3.2-2})}{2}=-\dfrac{4(\sqrt{4}+\sqrt{6-2})}{2}=\\\\\\\\=-\dfrac{4(2+\sqrt{4})}{2}=-2(2+2)=-2.4=-8$