lim (x3 + x2 + 3x + 3) / (x4 + x3 + 2x + 2)
x > -1
From the cave of the ancient stone tablets, we cleared away several feet of cobwebs and unearthed
"l'Hospital's" rule: If substitution of the limit results in ( 0/0 ), then the limit is equal to the
(limit of the derivative of the numerator) divided by (limit of the derivative of the denominator).
(3x2 + 2x + 3) / (4x3 + 3x2 + 2) evaluated at (x = -1) is:
(3 - 2 + 3) / (-4 + 3 + 2) = 4 / 1 = 1
2 plus 1-x divided by.
(-2x2 + x + 1) / (x - 1) = (-2x - 1)(x - 1)/(x - 1) = -2x - 1
Numerator = (x + 4)*(x + 3) Denominator = (2x - 1)*(x + 3) Calcelling out the common factor, (x + 3), the answer is (x + 4) divided by (2x - 1) provided x ≠0.5
Because all the terms have been divided by -1
(x2 + 2x + 15)/(x - 3) = x + 5 + 30/(x - 3) where x ≠3
9x/2x = 9/2 = 4.5
That depends on whether or not 2x is a plus or a minus
2 plus 1-x divided by.
(-2x2 + x + 1) / (x - 1) = (-2x - 1)(x - 1)/(x - 1) = -2x - 1
Numerator = (x + 4)*(x + 3) Denominator = (2x - 1)*(x + 3) Calcelling out the common factor, (x + 3), the answer is (x + 4) divided by (2x - 1) provided x ≠0.5
I assume you wish to simplify the expression (x - 2)/(3x + 9) . (2x + 6)/(2x - 4) = (x - 2)(2x + 6) / [(3x + 9)(2x - 4)] = 2(x - 2)(x + 3) / [6(x + 3)(x - 2)] = 1/3.
Because all the terms have been divided by -1
(x2 + 2x + 15)/(x - 3) = x + 5 + 30/(x - 3) where x ≠3
y = x / (x^2 + 2x + 1) The horizontal asymptote is y = 0
Assuming that he quadratic is 2x^2 + x - 15, the quotient is 2x - 5.
please elaborate your question
The quotient works out as: x^2+2x+4 and there is a remainder of -3