Ensino Superior ⇒ Limites no Infinito

[lucassmarques]

Calcule o limite:


[tex3]\lim_{x\rightarrow \infty }{\frac{\sqrt{t+\sqrt{t+\sqrt{t}}}}{\sqrt{t+1}}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\lim_{x\rightarrow \infty }{\frac{\sqrt{t+\sqrt{t+\sqrt{t}}}}{\sqrt{t+1}}}[/tex3]

[candre]

Código: Selecionar tudo

[tex3]\lim_{t\to+\infty}\frac{\sqrt{t+\sqrt{t+\sqrt{t}}}}{\sqrt{t+1}}=
\lim_{t\to+\infty}\sqrt{\frac{t+\sqrt{t+\sqrt{t}}}{t+1}}=
\sqrt{\lim_{t\to+\infty}\frac{t+\sqrt{t+\sqrt{t}}}{t+1}}=\\ \\
\sqrt{\lim_{t\to+\infty}\frac{\cancel{t}\left(1+\frac{\sqrt{t+\sqrt{t}}}{t}\right)}{\cancel{t}\left(1+\frac{1}{t}\right)}}=
\sqrt{\lim_{t\to+\infty}\frac{1+\frac{\sqrt{t+\sqrt{t}}}{t}}{1+\frac{1}{t}}}=
\sqrt{\lim_{t\to+\infty}\frac{1}{1+\cancelto{0}{\frac{1}{t}}}\cdot\lim_{t\to+\infty}1+\frac{\sqrt{t+\sqrt{t}}}{t}}=\\ \\
\sqrt{\lim_{t\to+\infty}1+\frac{\sqrt{t+\sqrt{t}}}{t}}=
\sqrt{\lim_{t\to+\infty}1+\lim_{t\to+\infty}\frac{\sqrt{t+\sqrt{t}}}{t}}=
\sqrt{\lim_{t\to+\infty}1+\lim_{t\to+\infty}\sqrt{\frac{t+\sqrt{t}}{t^2}}}=\\ \\
\sqrt{\lim_{t\to+\infty}1+\sqrt{\lim_{t\to+\infty}\frac{t+\sqrt{t}}{t^2}}}=
\sqrt{\lim_{t\to+\infty}1+\sqrt{\lim_{t\to+\infty}\frac{\cancel{t}}{t^{\cancel{2}}}+\frac{\sqrt{t}}{t^2}}}=
\sqrt{\lim_{t\to+\infty}1+\sqrt{\lim_{t\to+\infty}\frac{1}{t}+\frac{\sqrt{t}}{t^2}}}=\\ \\
\sqrt{\lim_{t\to+\infty}1+\sqrt{\lim_{t\to+\infty}\cancelto{0}{\frac{1}{t}}+\lim_{t\to+\infty}\frac{\sqrt{t}}{t^2}}}=
\sqrt{\lim_{t\to+\infty}1+\sqrt{\lim_{t\to+\infty}\frac{\sqrt{t}\sqrt{t}}{t^2\sqrt{t}}}}=
\sqrt{\lim_{t\to+\infty}1+\sqrt{\lim_{t\to+\infty}\frac{\cancel{t}}{t^\cancel{2}\sqrt{t}}}}}=\\ \\
\sqrt{1+\sqrt{\lim_{t\to+\infty}\cancelto{0}{\frac{1}{t\sqrt{t}}}}}=1[/tex3]

[RafaeldeLima]

Código: Selecionar tudo

[tex3]\lim_{t\to+\infty}\frac{\sqrt{t+\sqrt{t+\sqrt{t}}}}{\sqrt{t+1}}=\lim_{t\to+\infty}\sqrt{\frac{t+\sqrt{t+\sqrt{t}}}{t+1}}=\sqrt{\lim_{t\to+\infty}\frac{t+\sqrt{t+\sqrt{t}}}{t+1}}= \\ \\ \\ \\ \\ \sqrt{\lim_{t\to+\infty}\frac{\cancel{t}\left(1+\frac{\sqrt{t+\sqrt{t}}}{t}\right)}{\cancel{t}\left(1+\frac{1}{t}\right)}}=\sqrt{\lim_{t\to+\infty}\frac{1+\frac{\sqrt{t+\sqrt{t}}}{t}}{1+\frac{1}{t}}} = \sqrt{\lim_{t\to+\infty}\frac{1}{1+\frac{1}{t}}+\lim_{t\to+\infty}\frac{\frac{\sqrt{t+\sqrt{t}}}{t}}{1+\frac{1}{t}}} = \\ \\ \\ \\ \sqrt{\lim_{t\to+\infty}\frac{1}{1+\frac{1}{t}}+\lim_{t\to+\infty}\frac{\sqrt{\frac{t+\sqrt{t}}{t^2}}}{\frac{1}{1+\frac{1}{t}}}} = \sqrt{\lim_{t\to+\infty}\frac{1}{1+\frac{1}{t}}+\lim_{t\to+\infty}\frac{\sqrt{{\frac{1}{t}+\sqrt{\frac{t}{t^4}}}}}{\frac{1}{1+\frac{1}{t}}}} = \\ \\ \\ \\ \\ \sqrt{\lim_{t\to+\infty}\frac{1}{1+\frac{1}{t}}+\lim_{t\to+\infty}\frac{\sqrt{{\frac{1}{t}+\sqrt{\frac{1}{t^3}}}}}{\frac{1}{1+\frac{1}{t}}}} = \sqrt{\lim_{t\to+\infty}\frac{1}{1+\cancelto{0}{\frac{1}{t}}} \ \ +\lim_{t\to+\infty}\cancelto{0}{\frac{{\sqrt{{\cancelto{0}{\frac{1}{t}}+\sqrt{\cancelto{0}{\frac{1}{t^3}}}}}}}{\frac{1}{1+\cancelto{0}{\frac{1}{t}}}}}} = \\ \\ \\ \\ \boxed{\boxed{1}}[/tex3]