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- Politecnico di Milano
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- Energy Engineering
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- Analisi e geometria 2
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- Notes · By topic
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16 Limiti a due variabili
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LEZIONE16 12.4.2023 DEFINIZIONEDI LIMITE ~ f:D(=MM)-> I, jeGIRYp-todiaccumulazioneperD; Devopoterudiciamoche abricinare- ..2x lin f()=1=IR ce · e" punti- x-> 50 darmulaz*2x035= 0 SC5= 0(5,x))+C ⑳se OXd(5,5%=8 D NO- allora(f()- 2/15 BOLAt (alternativamate,.... Setaxo7520t.c. x =f(x)453nDalloraIf()- e/xs) Oss.1.Comeanalisitmeixla-ne estitutoda oxd(5,x) xa (ex0-8xxx x +d(nfx)ésotitutode x=H,(x+140]- Oss.2Comeinanalisia-limitese7UNICO. Inanalis1 existentahiI livefasex+ xtIlimf(n)e sono uguali2- 220- Inquestocaso abbiamomoltissimimodiper avvicinarciadi Itestatoiprindopername in FATTO:himf()= 1solose ffltende 5.->T allostesso Iqualsiasivia il mode in ainetendeadin D. In partistareinpuòtendereadilungoqualsiasicurva in Dr OSS-3Nowenecessariochenimf(x)=2-Anzif(a)potrebbewasexistere- Es-1-limyrD=B{(0,03(x,y)- (0,0) ↓Provinishingfrattem= olasosi0 None f(0,y)= -- 0-> 0 Iforma↑jr3F0 y= 0 hdiindecisione) quidlin.Seté=0 2 1Provorettay =e fare- ma- *= -Iafo x- 0 Abbiamotrovatochedirezionilungoqui Limiteéf,quindilim.7 GRAFICO LINEEd LIVELLOf(x,z) f(x,y) 082-f(x,y)ècostantelinerettede exce dell'orig ES- 2 f(x,y)= e D= RG(0,0] 10,0)podiccamulazioneberD. him seI- - (x,3)->(0,0) Perx= 0, y0 abbiamof(0,y)=0 - 0 y= 0 Pery= mx, xf(x,mx)- e s quindilim Se7i =0. -Intaleche S seIIC2,zill- vero Bastascegliere5(9)= 9_ Infetti(1,(y1=(n,y)11) Quindilim exy =0- (x,y)- (0,0)xyz GRAFICODIf(1,9)
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