Esercizio 6 
. Calcolare, se esiste, il seguente limite:
![\[\lim_{(x,y)\to (0,0)} \;\dfrac{x^4 + y^3}{x^2 + y^2}.\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-ab2369fdeb5f39878a6a493a44537539_l3.png "Rendered by QuickLaTeX.com")
Svolgimento. Sia 
tale che 
Passando in coordinate polari la funzione 
diventa
![\[f(\rho \cos \theta,\rho \sin \theta)=\tilde{f}(\rho,\theta)= \dfrac{\rho^4 \cos^4(\theta) + \rho^ 3 \sin^3(\theta)}{\rho^2} .\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-fc6fb908362bf0967568d1d8b4c3afff_l3.png "Rendered by QuickLaTeX.com")
Osserviamo che
![\[\left\lvert \tilde{f}(\rho,\theta) \right \lvert = \bigg \lvert \frac{\rho^2 (\rho^2 \cos^4(\theta) + \rho\sin^3(\theta))}{\rho^2} \bigg \lvert = |\rho^2 \cos^4(\theta) + \rho \sin^3(\theta)|.\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-c8f878e1ec6c4d83aea8d31b3778ca96_l3.png "Rendered by QuickLaTeX.com")
Per la disuguaglianza triangolare, e dato che 
e 
, abbiamo:
![\[|(\rho^2 \cos^4(\theta) + \rho \sin^3(\theta))| \leq |\rho^2 \cos^4(\theta) | + |\rho \sin^3(\theta)| \leq \rho^2 + \rho ,\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-7a7942abe553c90d48b6f1f540869e51_l3.png "Rendered by QuickLaTeX.com")
dove
![\[\lim_{\rho \rightarrow 0^+}(\rho^2 + \rho)=0.\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-42aee82c213c84c084a95d4d4e663b47_l3.png "Rendered by QuickLaTeX.com")
Si conclude che
![\[\boxcolorato{analisi}{ \lim_{(x,y) \to (0, 0)}\; \frac{x^4 y^3}{x^2 + y^2} = 0 .}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-f8286cf5da019bb75cc6c21c0b6a2834_l3.png "Rendered by QuickLaTeX.com")
Fonte: clicca qui.