Esercizio 41 limiti in due variabili

Limiti in due variabili

Home » Esercizio 41 limiti in due variabili

More results...

Generic selectors
Exact matches only
Search in title
Search in content
Post Type Selectors
post
page


 

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.