Esercizio 36 limiti in due variabili

Limiti in due variabili

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Esercizio 36 (\bigstar\largewhitestar\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)}{\left(x^2+y^2\right)^2}. \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

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

Si osserva che

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

dunque (1) diventa

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

Si conclude che

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