. Calcolare, se esiste, il seguente limite:
Svolgimento. Sia 
tale che 
Osserviamo che
![\[\begin{aligned} \left \vert xy \right \vert^{\left \vert xy \right \vert}-1&=e^{\left \vert xy \right \vert\ln \left \vert xy \right \vert}-1=1-1+\left \vert xy \right \vert\ln \left \vert xy \right \vert+o\left(\left \vert xy \right \vert\ln \left \vert xy \right \vert\right)=\\\\ &=\left \vert xy \right \vert\ln \left \vert xy \right \vert+o\left(\left \vert xy \right \vert\ln \left \vert xy \right \vert\right)\quad \text{per}\,\,(x,y)\rightarrow (0,0), \end{aligned}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-0ff10d4378c448c1fa549bbd71a9e950_l3.png "Rendered by QuickLaTeX.com")
da cui
![\[\begin{aligned}\label{2} \lim_{(x,y)\to (0,0)} \;\dfrac{\left \vert xy \right \vert^{\left \vert xy \right \vert}-1}{\sqrt{x^2+y^2}}&=\lim_{(x,y)\to (0,0)} \;\dfrac{\left \vert xy \right \vert\ln \left \vert xy \right \vert+o\left(\left \vert xy \right \vert\ln \left \vert xy \right \vert\right)}{\sqrt{x^2+y^2}}=\nonumber\\ &=\lim_{(x,y)\to (0,0)}\dfrac{\left \vert xy \right \vert \ln \left \vert xy \right \vert }{\sqrt{x^2+y^2}}\left(1+o\left(1\right)\right). \end{aligned}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-b34d1afdc1255a0f4209443771c9e0bd_l3.png "Rendered by QuickLaTeX.com")
Ora consideriamo
![\[g:\{(x,y)\in\mathbb{R}^2:\,xy\neq0\}\rightarrow \mathbb{R}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-d8f560d396b890be67fa008a896c15f9_l3.png "Rendered by QuickLaTeX.com")
tale che
![\[g(x,y)=\dfrac{\left \vert xy \right \vert \ln \left \vert xy \right \vert }{\sqrt{x^2+y^2}}.\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-5e283d3bc4ef2fe615a28a950fb79609_l3.png "Rendered by QuickLaTeX.com")
Passando le coordinate polari [1] ad 
, otteniamo:
![\[g(\rho \cos \theta,\rho \sin \theta)=\tilde{g}\left(\rho ,\theta\right)=\dfrac{\left \vert \dfrac{\rho^2\sin\left(2\theta\right)}{2}\right \vert \ln \left \vert \dfrac{\rho^2 \sin\left(2\theta\right)}{2}\right \vert }{\rho}=\left \vert \dfrac{\rho \sin\left(2\theta\right)}{2}\right \vert \ln \left \vert \frac{\rho^2 \sin\left(2\theta\right)}{2}\right \vert.\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-c8f6643bd5d6c597a031db6f2c3edd17_l3.png "Rendered by QuickLaTeX.com")
Osserviamo che [2]
![\[0\leq \left \vert \tilde{g}\left(\rho ,\theta\right)\right \vert \leq\left \vert\frac{\rho }{2} \ln \left \vert \frac{\rho^2}{2}\right \vert \right \vert ,\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-5c85fdf57c09b717656af67e53c6fd01_l3.png "Rendered by QuickLaTeX.com")
dove [3]
![\[\lim_{\rho \rightarrow 0^+}\left \vert \frac{\rho }{2} \ln \left \vert \frac{\rho^2}{2}\right \vert \right \vert =0.\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-efa9d801a7e2ac7967a3e9a584fc1ea6_l3.png "Rendered by QuickLaTeX.com")
Dunque, concludiamo che
![\[\boxcolorato{analisi}{\lim_{(x,y)\to (0,0)} \;\dfrac{\left \vert xy \right \vert^{\left \vert xy \right \vert}-1}{\sqrt{x^2+y^2}}=0.}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-30fd1021faf956275ee59f7442c0163a_l3.png "Rendered by QuickLaTeX.com")
1. Si ricorda che
![\[\begin{cases} x=\rho \cos \theta \\ \hspace{3.5cm} \rho\ge0,\,\theta \in [0,2\pi)\\ y=\rho \sin \theta \end{cases}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-6ec044fa3eafbf1ebec6e34727195737_l3.png "Rendered by QuickLaTeX.com")
2. 
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3. Il limite si dimostra che converge a zero, ad esempio, applicando il teorema di de l’Hôpital.
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