Esercizio 27 limiti in due variabili

Limiti in due variabili

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Esercizio 27  (\bigstar\bigstar\bigstar\largewhitestar\largewhitestar). Calcolare, se esiste, il seguente limite

(1)   \begin{equation*} \lim_{\left \vert \left \vert (x,y)\right \vert\right \vert\to +\infty } \dfrac{x^2+y^2}{x^4+y^4+xy}. \end{equation*}

 

Svolgimento. Sia f:\{(x,y)\in \mathbb{R}^2:\,x^4+y^4+xy\neq0\}\rightarrow \mathbb{R} tale che f(x,y)= \dfrac{x^2+y^2}{x^4+y^4+xy}.
Proviamo la restrizione y=0 di f, ottenendo

    \[f(x,0)=\dfrac{x^2}{x^4}=\dfrac{1}{x^2},\]

per cui (1) diventa

    \[\lim_{x \rightarrow +\infty}f(x,0)=\lim_{x \rightarrow+ \infty}\dfrac{1}{x^2}=0.\]

Ora proviamo la restrizione x=0 avendo così

    \[f(0,y)=\dfrac{y^2}{y^4}=\dfrac{1}{y^2}\]

ed (1) diventa

    \[\lim_{y \rightarrow +\infty}f(0,y)=\lim_{y \rightarrow +\infty}\dfrac{1}{y^2}=0.\]

Proviamo a dimostrare che (1) converge a zero riscrivendo f in coordinate polari

Dunque si ha

    \[\begin{aligned} f(\rho \cos \theta ,\rho \sin \theta)&=\tilde{f}\left(\rho,\theta\right)=\dfrac{\rho^2}{\rho^4\left(\cos^4\theta+\sin^4\theta\right)+\rho^2\cos \theta \sin \theta}=\dfrac{\rho^2}{\rho^4\left(\cos^4\theta+\sin^4\theta\right)+\dfrac{1}{2}\rho^2\sin\left(2\theta\right)}=\\ &=\dfrac{1}{\rho^2\left(\cos^4\theta+\sin^4\theta\right)+\dfrac{1}{2}\sin\left(2\theta\right)}=\dfrac{1}{\rho^2\left(\left(\cos^2\theta+\sin^2\theta\right)^2-2\cos^2\theta \sin^2\theta\right)+\dfrac{1}{2}\sin\left(2\theta\right)}=\\ &=\dfrac{1}{\rho^2-2\rho^2\cos^2\theta\sin^2\theta+\dfrac{1}{2}\sin\left(2\theta\right)}=\dfrac{1}{\rho^2-\dfrac{1}{2}\rho^2\sin^2\left(2\theta\right)+\dfrac{1}{2}\sin\left(2\theta\right)}. \end{aligned}\]

Osserviamo che

    \[-1\leq \sin\left(2\theta\right)\leq 1\]

e

    \[-1\leq -6\sin^2\left(2\theta\right)\leq 0,\]

per ogni \theta \in (0,2\pi). Pertanto

    \[\rho^2-\dfrac{\rho^2}{2}-\dfrac{1}{2}\leq \rho^2-\dfrac{1}{2}\rho^2\sin^2\left(2\theta\right)+\dfrac{1}{2}\sin\left(2\theta\right)\leq \rho^2+\dfrac{1}{2},\]

cioè

    \[\dfrac{\rho^2}{2}-\dfrac{1}{2}\leq \rho^2-\dfrac{1}{2}\rho^2\sin^2\left(2\theta\right)+\dfrac{1}{2}\sin\left(2\theta\right)\leq \rho^2+\dfrac{1}{2},\]

da cui, per \rho>1, si ha

    \[\dfrac{1}{ \rho^2+\dfrac{1}{2}}\leq \tilde{f}\left(\rho,\theta\right)\leq \dfrac{1}{\dfrac{\rho^2}{2}-\dfrac{1}{2}}.\]

Inoltre, si ha

    \[\lim_{\rho\to+\infty}\dfrac{1}{ \rho^2+\dfrac{1}{2}}=\lim_{\rho\to+\infty}\dfrac{1}{\dfrac{\rho^2}{2}-\dfrac{1}{2}}=0.\]

Quindi per il teorema del confronto si ha

    \[\lim_{\rho\to+\infty}\tilde{f}\left(\rho,\theta\right)=0.\]

Si conclude che

    \[\boxcolorato{analisi}{\lim_{\left \vert \left \vert (x,y)\right \vert\right \vert\to +\infty } \dfrac{x^2+y^2}{x^4+y^4+xy}=0.}\]

 

Fonte: clicca qui.