Question

Determine o limite lim x tende -2= raiz 5-raiz7+y/raizy2+3-raiz7...pelo amor de Deus

Answer 1

Dado o limite:

$\mathsf{\lim _{y\to -2\:}\left(\dfrac{\sqrt{5}-\sqrt{7+y}}{\sqrt{y^2+3}-\sqrt{7}}\right)}$

Observa-se que não podemos simplesmente substituir -2 no lugar de y, pois daria uma indeterminação:

$\mathsf{\lim _{y\to -2}\left(\dfrac{\sqrt{5}-\sqrt{7+y}}{\sqrt{y^2+3}-\sqrt{7}}\right)=\dfrac{\sqrt{5}-\sqrt{5}}{\sqrt{7}-\sqrt{7}}=\dfrac{0}{0}}$

= = = = =

Em caso de uma indeterminação 0/0 ou ∞/∞, usaremos a regra de L' Hopital:

$\boxed{\mathsf{\lim _{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=\lim _{x\to a}\frac{f'\left(x\right)}{g'\left(x\right)}}}$

= = = = =

Derivando o numerador: $\mathsf{f\left(x\right)=\sqrt{5}-\sqrt{7+y}}$

$\mathsf{=\dfrac{d}{dx}\left(\sqrt{5}-\sqrt{7+y}\right)}\\\\\\\mathsf{=\dfrac{d}{dx}\left(\sqrt{5}\right)-\dfrac{d}{dx}\left(\sqrt{7+y}\right)}\\\\\\\mathsf{=0-\dfrac{1}{2\sqrt{7+y}}}\\\\\\\mathsf{=-\dfrac{1}{2\sqrt{7+y}}}$

= = = = =

Derivando o denominador: $\mathsf{g\left(x\right)=\sqrt{y^2+3}-\sqrt{7}}$

$\mathsf{=\dfrac{d}{dx}\left(\sqrt{y^2+3}-\sqrt{7}\right)}\\\\\\\mathsf{=\dfrac{d}{dx}\left(\sqrt{y^2+3}\right)-\dfrac{d}{dx}\left(\sqrt{7}\right)}\\\\\\\mathsf{=\dfrac{x}{\sqrt{x^2+3}}-0}\\\\\\\mathsf{=\dfrac{x}{\sqrt{x^2+3}}}$

= = = = =

Logo:

$\mathsf{=\lim _{y\to -2}\left(\dfrac{\sqrt{5}-\sqrt{7+y}}{\sqrt{y^2+3}-\sqrt{7}}\right)}\\\\\\\mathsf{=\lim _{y\to -2}\left(\dfrac{-\frac{1}{2\sqrt{7+y}}}{\frac{x}{\sqrt{x^2+3}}}\right)}\\\\\\\mathsf{=\lim _{y\to -2}\left(-\dfrac{1}{2\sqrt{7+y}}\cdot \dfrac{\sqrt{x^2+3}}{x}\right)}\\\\\\\mathsf{=\lim _{y\to -2}\left(-\dfrac{\sqrt{x^2+3}}{2x\sqrt{7+y}}\right)}$

Substituindo -2 no lugar de $\mathsf{y}$ acharemos o valor do limite:

$\mathsf{=-\dfrac{\sqrt{\left(-2\right)^2+3}}{2\cdot \left(-2\right)\sqrt{7+\left(-2\right)}}}\\\\\\\mathsf{=-\dfrac{\sqrt{4+3}}{-4\sqrt{7-2}}}\\\\\\\mathsf{=\dfrac{\sqrt{7}}{4\sqrt{5}}}\\\\\\\mathsf{=\dfrac{\sqrt{7}}{4\sqrt{5}}\cdot \dfrac{\sqrt{5}}{\sqrt{5}}}\\\\\\\mathsf{=\dfrac{\sqrt{35}}{4\cdot 5}}\\\\\\\mathsf{=\boxed{\mathsf{\dfrac{\sqrt{35}}{20}}}}\: \: \checkmark$