Esercizio 37 limiti in due variabili

Limiti in due variabili

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Esercizio 37  (\bigstar\bigstar\largewhitestar\largewhitestar\largewhitestar). Calcolare, se esiste, il seguente limite

(1)   \begin{equation*} \lim_{(x,y)\to (0,0)} \;\dfrac{\ln\left(\cos\left(x^2+y^2\right)\right)+\dfrac{1}{2}\left(x^2+y^2\right)^2}{\left(x^2+y^2\right)^4}. \end{equation*}

 

 

Svolgimento.  Sia

    \[f:\Omega=\{(x,y)\in\mathbb{R}^2:\,\cos\left(x^2+y^2\right)>0,\,(x,y)\neq (0,0) \}\rightarrow \mathbb{R}\]

tale che

    \[f(x,y)=\dfrac{\ln\left(\cos\left(x^2+y^2\right)\right)+\dfrac{1}{2}\left(x^2+y^2\right)^2}{\left(x^2+y^2\right)^4}.\]

Si osserva che

    \[\begin{aligned} \ln\left(\cos\left(x^2+y^2\right)\right)&=\ln\left(1-\dfrac{1}{2}\left(x^2+y^2\right)^2+\dfrac{1}{24}\left(x^2+y^2\right)^4+o\left(\left(x^2+y^2\right)^4\right)\right)=\\\\ &=-\dfrac{1}{2}\left(x^2+y^2\right)^2+\dfrac{1}{24}\left(x^2+y^2\right)^4-\dfrac{1}{2}\cdot\dfrac{1}{4}\left(x^2+y^2\right)^4 +o\left(\left(x^2+y^2\right)^4\right)=\\\\ & -\dfrac{1}{2}\left(x^2+y^2\right)^2-\dfrac{1}{12}\left(x^2+y^2\right)^4+o\left(\left(x^2+y^2\right)^4\right)\quad\text{per}\,\,(x,y)\rightarrow(0,0), \end{aligned}\]

dunque (1) diventa

    \[\begin{aligned} & \lim_{(x,y)\to (0,0)} \;\dfrac{\ln\left(\cos\left(x^2+y^2\right)\right)+\dfrac{1}{2}\left(x^2+y^2\right)^2}{\left(x^2+y^2\right)^4}=\\\\ &= \lim_{(x,y)\to (0,0)} \;\dfrac{-\dfrac{1}{2}\left(x^2+y^2\right)^2-\dfrac{1}{12}\left(x^2+y^2\right)^4+\dfrac{1}{2}\left(x^2+y^2\right)^2+o\left(\left(x^2+y^2\right)^4\right)}{\left(x^2+y^2\right)^4}=\\\\ &= \lim_{(x,y)\to (0,0)} \;\dfrac{-\dfrac{1}{12}\left(x^2+y^2\right)^4+o\left(\left(x^2+y^2\right)^4\right)}{\left(x^2+y^2\right)^4}=-\dfrac{1}{12}. \end{aligned}\]

Si conclude che

    \[\boxcolorato{analisi}{\lim_{(x,y)\to (0,0)} \;\dfrac{\ln\left(\cos\left(x^2+y^2\right)\right)+\dfrac{1}{2}\left(x^2+y^2\right)^2}{\left(x^2+y^2\right)^4}=-\dfrac{1}{12}.}\]