Esercizio 41 limiti in due variabili

Limiti in due variabili

Home » Esercizio 41 limiti in due variabili


 

Esercizio 41   (\bigstar\bigstar\bigstar\bigstar\largewhitestar). Calcolare, se esiste, il seguente limite:

(1)   \begin{equation*} \lim_{(x,y)\to (0,0)} \;\dfrac{\left \vert xy \right \vert^{\left \vert xy \right \vert}-1}{\sqrt{x^2+y^2}}. \end{equation*}

 

Svolgimento.  Sia f:\{(x,y)\in\mathbb{R}^2:\,xy\neq0\}\rightarrow \mathbb{R} tale che f(x,y)=\dfrac{\left \vert xy \right \vert^{\left \vert xy \right \vert}-1}{\sqrt{x^2+y^2}}.
Osserviamo che

    \[\begin{aligned} \left \vert xy \right \vert^{\left \vert xy \right \vert}-1&=e^{\left \vert xy \right \vert\ln \left \vert xy \right \vert}-1=1-1+\left \vert xy \right \vert\ln \left \vert xy \right \vert+o\left(\left \vert xy \right \vert\ln \left \vert xy \right \vert\right)=\\\\ &=\left \vert xy \right \vert\ln \left \vert xy \right \vert+o\left(\left \vert xy \right \vert\ln \left \vert xy \right \vert\right)\quad \text{per}\,\,(x,y)\rightarrow (0,0), \end{aligned}\]

da cui

    \[\begin{aligned}\label{2} \lim_{(x,y)\to (0,0)} \;\dfrac{\left \vert xy \right \vert^{\left \vert xy \right \vert}-1}{\sqrt{x^2+y^2}}&=\lim_{(x,y)\to (0,0)} \;\dfrac{\left \vert xy \right \vert\ln \left \vert xy \right \vert+o\left(\left \vert xy \right \vert\ln \left \vert xy \right \vert\right)}{\sqrt{x^2+y^2}}=\nonumber\\ &=\lim_{(x,y)\to (0,0)}\dfrac{\left \vert xy \right \vert \ln \left \vert xy \right \vert }{\sqrt{x^2+y^2}}\left(1+o\left(1\right)\right). \end{aligned}\]

Ora consideriamo

    \[g:\{(x,y)\in\mathbb{R}^2:\,xy\neq0\}\rightarrow \mathbb{R}\]

tale che

    \[g(x,y)=\dfrac{\left \vert xy \right \vert \ln \left \vert xy \right \vert }{\sqrt{x^2+y^2}}.\]

Passando le coordinate polari [1] ad g, otteniamo:

    \[g(\rho \cos \theta,\rho \sin \theta)=\tilde{g}\left(\rho ,\theta\right)=\dfrac{\left \vert \dfrac{\rho^2\sin\left(2\theta\right)}{2}\right \vert \ln \left \vert \dfrac{\rho^2 \sin\left(2\theta\right)}{2}\right \vert }{\rho}=\left \vert \dfrac{\rho \sin\left(2\theta\right)}{2}\right \vert \ln \left \vert \frac{\rho^2 \sin\left(2\theta\right)}{2}\right \vert.\]

Osserviamo che [2]

    \[0\leq \left \vert \tilde{g}\left(\rho ,\theta\right)\right \vert \leq\left \vert\frac{\rho }{2} \ln \left \vert \frac{\rho^2}{2}\right \vert \right \vert ,\]

dove [3]

    \[\lim_{\rho \rightarrow 0^+}\left \vert \frac{\rho }{2} \ln \left \vert \frac{\rho^2}{2}\right \vert \right \vert =0.\]

Dunque, concludiamo che

    \[\boxcolorato{analisi}{\lim_{(x,y)\to (0,0)} \;\dfrac{\left \vert xy \right \vert^{\left \vert xy \right \vert}-1}{\sqrt{x^2+y^2}}=0.}\]

 

1. Si ricorda che

    \[\begin{cases} x=\rho \cos \theta \\ \hspace{3.5cm} \rho\ge0,\,\theta \in [0,2\pi)\\ y=\rho \sin \theta \end{cases}\]

2. \left \vert \sin\left(2\theta\right)\right \vert \leq1.

3. Il limite si dimostra che converge a zero, ad esempio, applicando il teorema di de l’Hôpital.