Esercizio 38 limiti in due variabili

Limiti in due variabili

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Esercizio 38 (\bigstar\bigstar\largewhitestar\largewhitestar\largewhitestar). Calcolare, se esiste, il seguente limite

(1)   \begin{equation*} \lim_{(x,y)\to (0,0)} \;\dfrac{\left(e^{x^2+y^3}-1-\left(x^2+y^3\right)\right)\left(1+\cos\left(x^2+y^3\right)\right)}{\sin^2\left(x^2+y^3\right)+\tan^2\left(x^2+y^3\right)}. \end{equation*}

 

 

Svolgimento. Sia

    \[f:\{(x,y)\in\mathbb{R}^2:\,\sin^2\left(x^2+y^3\right)+\tan^2\left(x^2+y^3\right)\neq 0 \}\rightarrow \mathbb{R}\]

tale che

    \[f(x,y)=\dfrac{\left(e^{x^2+y^3}-1-\left(x^2+y^3\right)\right)\left(1+\cos\left(x^2+y^3\right)\right)}{\sin^2\left(x^2+y^3\right)+\tan^2\left(x^2+y^3\right)}.\]

Posto x^2+y^3=t, il limite diventa

    \[\lim_{t \rightarrow 0 }\dfrac{\left(e^t-1-t\right)\left(1+\cos t\right)}{\sin^2t+\tan^2 t}.\]

Per t \rightarrow 0 si hanno i seguenti sviluppi:

    \[\begin{aligned} &a) \, e^t-1-t=1+t+\dfrac{t^2}{2}-1-t+o\left(t^2\right)=\dfrac{t^2}{2}+o\left(t^2\right);\\\\ & b)\, 1+\cos t =1+1+o\left(1\right)=2+o\left(1\right);\\\\ &c)\, \sin^2t + \tan^2 t =t^2+t^2+o\left(t^2\right)=2t^2+o\left(t^2\right). \end{aligned}\]

Dunque la (1) diventa

    \[\begin{aligned} &\lim_{t \rightarrow 0}\dfrac{\left(e^t-1-t\right)\left(1+\cos t\right)}{\sin^2t+\tan^2 t}=\lim_{t \rightarrow 0}\dfrac{\left(\frac{t^2}{2}+o\left(t^2\right)\right)\left(2+o\left(1\right)\right)}{2t^2+o\left(t^2\right)}=\dfrac{1}{2}. \end{aligned}\]

Si conclude che

    \[\boxcolorato{analisi}{\lim_{(x,y)\to (0,0)} \;\dfrac{\left(e^{x^2+y^3}-1-\left(x^2+y^3\right)\right)\left(1+\cos\left(x^2+y^3\right)\right)}{\sin^2\left(x^2+y^3\right)+\tan^2\left(x^2+y^3\right)}=\dfrac{1}{2}.}\]