Question

Resolva: $\lim_{x\to 1 }\frac{\sqrt{x}-1}{\sqrt{2x+3}-\sqrt{5}}$

Answer 1

Explicação passo-a-passo:

$\sf lim_{x\to~1}~\dfrac{\sqrt{x}-1}{\sqrt{2x+3}-\sqrt{5}}$

$\sf =lim_{x\to~1}~\dfrac{\sqrt{x}-1}{\sqrt{2x+3}-\sqrt{5}}\cdot\dfrac{\sqrt{2x+3}+\sqrt{5}}{\sqrt{2x+3}+\sqrt{5}}$

$\sf =lim_{x\to~1}~\dfrac{(\sqrt{x}-1)\cdot(\sqrt{2x+3}+\sqrt{5})}{(\sqrt{2x+3})^2-(\sqrt{5})^2}$

$\sf =lim_{x\to~1}~\dfrac{(\sqrt{x}-1)\cdot(\sqrt{2x+3}+\sqrt{5})}{2x+3-5}$

$\sf =lim_{x\to~1}~\dfrac{(\sqrt{x}-1)\cdot(\sqrt{2x+3}+\sqrt{5})}{2x-2}$

$\sf =lim_{x\to~1}~\dfrac{(\sqrt{x}-1)\cdot(\sqrt{2x+3}+\sqrt{5})}{2\cdot(x-1)}$

$\sf =lim_{x\to~1}~\dfrac{(\sqrt{x}-1)\cdot(\sqrt{2x+3}+\sqrt{5})}{2\cdot(x-1)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+1}$

$\sf =lim_{x\to~1}~\dfrac{[(\sqrt{x})^2-1^2]\cdot(\sqrt{2x+3}+\sqrt{5})}{2\cdot(x-1)\cdot(\sqrt{x}+1)}$

$\sf =lim_{x\to~1}~\dfrac{(x-1)\cdot(\sqrt{2x+3}+\sqrt{5})}{2\cdot(x-1)\cdot(\sqrt{x}+1)}$

$\sf =lim_{x\to~1}~\dfrac{\sqrt{2x+3}+\sqrt{5}}{2\cdot(\sqrt{x}+1)}$

$\sf =\dfrac{\sqrt{2\cdot1+3}+\sqrt{5}}{2\cdot(\sqrt{1}+1)}$

$\sf =\dfrac{\sqrt{2+3}+\sqrt{5}}{2\cdot(1+1)}$

$\sf =\dfrac{\sqrt{5}+\sqrt{5}}{2\cdot2}$

$\sf =\dfrac{2\sqrt{5}}{4}$

$\sf =\dfrac{\sqrt{5}}{2}$