Question

calcule o limite raiz 5+y - raiz 5 / y quando y=0

Answer 1

$\lim\limits_{y\rightarrow0}\dfrac{\sqrt{5+y}-\sqrt{5}}{y}$

Multiplicando o numerador e o denominador pelo conjugado do numerador:

$\lim\limits_{y\rightarrow0}\dfrac{\sqrt{5+y}-\sqrt{5}}{y}=\lim\limits_{y\rightarrow0}\dfrac{(\sqrt{5+y}-\sqrt{5})\cdot(\sqrt{5+y}+\sqrt{5})}{y\cdot(\sqrt{5+y}+\sqrt{5})}\\\\\\\lim\limits_{y\rightarrow0}\dfrac{\sqrt{5+y}-\sqrt{5}}{y}=\lim\limits_{y\rightarrow0}\dfrac{(\sqrt{5+y})^{2}-(\sqrt{5})^{2}}{y(\sqrt{5+y}+\sqrt{5})}\\\\\\\lim\limits_{y\rightarrow0}\dfrac{\sqrt{5+y}-\sqrt{5}}{y}=\lim\limits_{y\rightarrow0}\dfrac{5+y-5}{y(\sqrt{5+y}+\sqrt{5})}=\lim\limits_{y\rightarrow0}\dfrac{y}{y(\sqrt{5+y}+\sqrt{5})}$

Cortando y:

$\lim\limits_{y\rightarrow0}\dfrac{\sqrt{5+y}-\sqrt{5}}{y}=\lim\limits_{y\rightarrow0}\dfrac{1}{\sqrt{5+y}+\sqrt{5}}\\\\\\\lim\limits_{y\rightarrow0}\dfrac{\sqrt{5+y}-\sqrt{5}}{y}=\dfrac{1}{\sqrt{5+0}+\sqrt{5}}\\\\\\\lim\limits_{y\rightarrow0}\dfrac{\sqrt{5+y}-\sqrt{5}}{y}=\dfrac{1}{\sqrt{5}+\sqrt{5}}\\\\\\\boxed{\boxed{\lim\limits_{y\rightarrow0}\dfrac{\sqrt{5+y}-\sqrt{5}}{y}=\dfrac{1}{2\sqrt{5}}}}$

Se quiser racionalizar:

$\boxed{\boxed{\lim\limits_{y\rightarrow0}\dfrac{\sqrt{5+y}-\sqrt{5}}{y}=\dfrac{\sqrt{5}}{10}}}$

Answer 2

$\lim\limits_{y\to0}\dfrac{\sqrt{5+y}-\sqrt{5}}{y}=\lim\limits_{y\to0}\left(\dfrac{\sqrt{5+y}-\sqrt{5}}{y}\times\dfrac{\sqrt{5+y}+\sqrt{5}}{\sqrt{5+y}+\sqrt{5}}\right)=\\\\\\=\lim\limits_{y\to0}\dfrac{5+y+\sqrt{25+5y}-\sqrt{25+5y}-5}{y(\sqrt{5+y}+\sqrt{5})}=\lim\limits_{y\to0}\dfrac{y}{y(\sqrt{5+y}+\sqrt{5})}=\\\\\\=\lim\limits_{y\to0}\dfrac{1}{\sqrt{5+y}+\sqrt{5}}=\dfrac{1}{\sqrt{5+0}+\sqrt{5}}=\dfrac{1}{2\sqrt{5}}=\bold{\dfrac{\sqrt{5}}{10}}$