lim 4x3- 2x2 + x 4x3+2x

Giải:

Ta có:

lim 4x3- 2x2 + x 4x3+2x = lim (4x3- 2x2 + x) - lim (4x3+2x) = lim (2x2-2x) = lim 2x2 - lim 2x = lim 2x(x-1) = 2(lim x(x-1)) = 2(lim x2-lim x) = 2(0-0) = 0

Vậy, lim 4x3- 2x2 + x 4x3+2x = 0.

lim r? – 52 +6 1 x-2 22 12x+20

Giải:

Ta có:

lim r? – 52 +6 1 x-2 22 12x+20 = lim (r2-52+6) - lim (12x+20)/22 = lim (r2-25+6) - lim (6x+10)/4 = lim (r-5)2 - lim (3x+5)/2 = (lim r-lim 5)2 - lim (3x+5)/2 = (lim r)2 - lim (3x+5)/2 = (-5)2 - lim (3x+5)/2 = 25 - lim (3x+5)/2 = 25 - (3lim x+5)/2 = 25 - (3(lim x)+5)/2 = 25 - (3(-3)+5)/2 = 25 - (-4)/2 = 25 + 2 = 27

Vậy, lim r? – 52 +6 1 x-2 22 12x+20 = 27.

lim √2x+1-3 →4√x-2-√2

Giải:

Ta có:

lim √2x+1-3 →4√x-2-√2 = lim (√2x+1-3) - lim (4√x-2-√2) = lim √2x+1 - lim 3 - lim 4√x + lim 2 - lim √2 = lim (√2x+1) - 3 - lim 4√x + 2 - lim √2 = lim (2√x/2) - 3 - lim (2√x) + 2 - lim √2 = lim √x - 3 - lim 2√x + 2 - lim √2 = (lim x)^(1/2) - 3 - 2(lim x)^(1/2) + 2 - lim √2 = (-3)^(1/2) - 3 - 2(-3)^(1/2) + 2 - lim √2 = 0 - 3 - 6 + 2 - lim √2 = -9 - lim √2 = -9 - √2

Kết luận:

lim √2x+1-3 →4√x-2-√2 = -9 - √2, khi x tiến đến -3.

lim VI-1

Giải:

Ta có:

lim VI-1 = lim (x-1)^(1/2) = (lim x

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