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Oppgave d)

lim x-> -1 2x^2+10x+8/3x+3=lim x-> -1 (x+1)(x+4)/3(x+1)=lim x->-1 x+4/3=lim x->-1 (1/3x+4/3)=1/3*-1+4/3=1

[tex]\lim_{x \rightarrow -1} \ \frac{2x^2 + 10x + 8}{3x+3}[/tex]

[tex]\lim_{x \rightarrow -1} \ \frac{2(x+1)(x+4)}{3(x+1)}[/tex]

Der har du feilen din.

[tex]ax^2 + bx + c = a(x-x_1)(x-x_2)[/tex]

Da blir det [tex]\lim_{x\rightarrow -1}\ \frac {2(x+4)}{3}=\frac {2(-1+4)}{3}=\frac {-2+8}{3}=\frac {6}{3}=2[/tex]


Nå begynner det å komme,thanks..