Question

$\lim_{n \to \infty} \frac{\sqrt{n} }{\sqrt{n+\sqrt{n+\sqrt{n} } } }$

Como resolvo este exercício, é sobre limite.

Answer 1

$\large\boxed{\begin{array}{l}\displaystyle\rm\lim_{n \to \infty}\dfrac{\sqrt{n}}{\sqrt{n+\sqrt{n+\sqrt{n}}}}=\sqrt{\lim_{n \to \infty}\dfrac{n}{n+\sqrt{n+\sqrt{n}}}}\\\\\displaystyle\rm\sqrt{\lim_{n \to \infty}\dfrac{\frac{1}{\backslash\!\!\!n}\cdot \backslash\!\!\!n}{\frac{1}{n}\cdot(n+\sqrt{n+\sqrt{n}})}}\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\rm\sqrt{\lim_{n \to\infty}\dfrac{1}{1+\dfrac{1}{n}\sqrt{n+\sqrt{n}}}}=\sqrt{\lim_{n \to \infty}\dfrac{1}{1+\sqrt{\bigg(\dfrac{1}{n}\bigg)^2\cdot(n+\sqrt{n})}}}\\\\\displaystyle\rm\sqrt{\lim_{n \to \infty}\dfrac{1}{1+\sqrt{\dfrac{1}{n^2}(n+\sqrt{n})}}}\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\rm\sqrt{\lim_{n \to\infty}\dfrac{1}{1+\sqrt{\dfrac{1}{n}+\dfrac{1}{n^2}\sqrt{n}}}}\\\\\displaystyle\rm\sqrt{\lim_{n \to \infty}\dfrac{1}{1+\sqrt{\dfrac{1}{n}+\sqrt{\dfrac{1}{\diagup\!\!\!\!\!n^4}\cdot \diagup\!\!\!\!n}}}}\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\rm\sqrt{\lim_{n\to \infty}\dfrac{1}{1+\sqrt{\dfrac{1}{n}+\sqrt{\dfrac{1}{n^3}}}}}\\\sf substituindo\,temos:\\\rm\sqrt{\dfrac{1}{1+\sqrt{0+\sqrt{0}}}}=\sqrt{\dfrac{1}{1}}=\sqrt{1}=1\end{array}}$

$\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\rm\lim_{n \to \infty}\dfrac{\sqrt{n}}{\sqrt{n+\sqrt{n+\sqrt{n}}}}=1}}}}\checkmark$