Limite

[angelox9]

Salve a tutti, ho il seguente limite:

\(\displaystyle \lim_{x \to +\infty } arctan\left(\frac{3x^3+2x^2}{x^3}\right)\frac{1-cos\left(\frac{1}{\sqrt[3]{x}+1}\right)}{sin^2\frac{1}{\sqrt{x}+2}} = \)

\(\displaystyle \lim_{x \to +\infty } arctan\left(\frac{3x^3(1+\frac{2}{3x})}{x^3}\right)\frac{1-cos\left(\frac{1}{\sqrt[3]{x}+1}\right)}
{\left(\frac{1}{\sqrt[3]{x}+1}\right)^2}
\frac{\left(\frac{1}{\sqrt{x}+2}\right)^2}
{sin^2\frac{1}{\sqrt{x}+2}}
\frac{\left(\frac{1}{\sqrt[3]{x}+1}\right)^2}{\left(\frac{1}{\sqrt{x}+2}\right)^2}= \)

\(\displaystyle \lim_{x \to +\infty } arctan\left[3(1+\frac{2}{3x})\right]\frac{1-cos\left(\frac{1}{\sqrt[3]{x}+1}\right)}
{\left(\frac{1}{\sqrt[3]{x}+1}\right)^2}
\frac{\left(\frac{1}{\sqrt{x}+2}\right)^2}
{sin^2\frac{1}{\sqrt{x}+2}}
\left(\frac{\sqrt{x}+2}{\sqrt[3]{x}+1}\right)^2 = \)

\(\displaystyle \lim_{x \to +\infty } arctan\left[3(1+\frac{2}{3x})\right]\frac{1-cos\left(\frac{1}{\sqrt[3]{x}+1}\right)}
{\left(\frac{1}{\sqrt[3]{x}+1}\right)^2}
\frac{\left(\frac{1}{\sqrt{x}+2}\right)^2}
{sin^2\frac{1}{\sqrt{x}+2}}
\frac{x+4\sqrt{x}+4}{\left(\sqrt[3]{x}+1\right)^2} = +\infty\)

La prima part del limite fa: \(\displaystyle arctan (3) \)
La seconda parte del limite fa: \(\displaystyle \frac{1}{2} \)
La terza parte del limite fa: \(\displaystyle 1 \)
La quarta parte del limite fa: \(\displaystyle +\infty \)
Il risultato finale fa: \(\displaystyle +\infty \)

Giusto?

@pilloeffe

[Antimius]

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