Question

Urgent. sa se calculexe:
a) lim unde x→2 din (x²+3x- 10)/ (2x²-x-6 )
b) lim unde x→1 din (√x- 1 )/ (x²-1 )
l) lim unde x→-1 din (∛x +1)/ (x³+ 1)

Answer 1

1.x²+3x-10=0
x1=-5  x2=2  Ecuatia  devine
(x+5)(x-2)=0
2x²-x-6=0  x1=- 3/2  x2=2
(x+3/2)(x-2)=0
lim (x+5)(x-2)/(x+3/2)(x-2)=        x→2
lim(x+5)/(x+3/2)=2
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2.Prelucram  fractia
(√x-1)/(x²-1)=(√x-1)/(x-1)(x+1)=(√x-1)/(√x-1)(√x+1)(x+1)        x→1
lim1/(√x+1)(x+1)=1/4
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3.Aplici  formula  A³+B³=(A²-AB+B²)
prelucrezi  numitorul
x³+1=(x+1)(x²-x+1)=(∛x+1)(∛x²-∛x+1)(x²-x+1)
x→ -1
lim(∛x+1)/(∛x+1)(∛x²-∛x+1)(x²+x+1)=lim1/(∛x²-∛x+1)(x²-x+1)=1/3·3=1/9