Limx tends to infinity 2x ^ 2-4 / 3x ^ 2-x

Limx tends to infinity 2x ^ 2-4 / 3x ^ 2-x

Limx tends to be infinite 2x ^ 2-4 / 3x ^ 2-x. the numerator denominator is divided by X & # 178;
=Limx tends to infinity (2-4 / x ^ 2) / (3-1 / x)
=2/3

The limit limx tends to 0 in (1 + 2x) / 3x,

Lobita's law, type 0 / 0, in (1 + 2x) / 3x, upper and lower derivatives, 2 / 3

Limx --- 0f (x) / x = a, where f (0) = 0 and the derivative exists,

A is the slope of F (x) at x = 0
A=lim f(x)-f(0) / x-0

Let f (x) be continuous on [0,3] and differentiable in (0,3), and f (0) + F (1) + F (2) = 3, f (3) = 1. It is proved that there must be ξ ∈ (0,3) such that f ′ (ξ) = 0

Because f (x) is continuous on [0,3], so f (x) is continuous on [0,2], and there must be a maximum m and a minimum m on [0,2], so: m ≤ f (0) ≤ m, m ≤ f (1) ≤ m, m ≤ f (2) ≤ m, so: m ≤ f (0) + F (1) + F (2) 3 ≤ M

Let the function be continuous on f (x) and differentiable in (1,0). It is proved that there is at least one point ξ ∈ (0,1) such that f '(ξ) = 2 ξ [f (1) - f (0)]

Constructor g (x) = f (x) - (the square of x) [f (1) - f (0)]
G(1)=f(1)-[f(1)-f(0)]=f(0)
G (0) = f (0) - 0 = f (0) known from Cauchy mean value theorem
There is a point ξ such that G '(ξ) = 0
G'(x )=f'(x )-2x[f(1)-f(0)]
G '(ξ) = f' (ξ) - 2 ξ [f (1) - f (0)] = 0, that is, there is a point ξ such that f '(ξ) = 2 ξ [f (1) - f (0)]

Let f (x) be continuous on a closed interval [0, a] and f (0) = 0. If f '(x) exists and is an increasing function (0)

Let f '(x) = [XF' (x) - f (x)] / x ^ 2, let g (x) = [XF '(x) - f (x)]' = XF '' (x). Since the derivative of F (x) on [0, a] exists and is an increasing function, the second derivative of F (x) on [0, a] is greater than 0, so g (x) is greater than 0, f (x) = f (x) / X is an increasing function

Limx - > 1 (x ^ 3 + ax ^ 2 + X + b) / (x ^ 2-1) = 3 to find the value of a and B

When x tends to 1, substituting 1 into the numerator denominator, we can see that the numerator is a + B + 2, and the denominator is 0. Because of the existence of limit, we get a + B + 2 = 0. Then we use the law of lobita to seek the upper and lower derivatives. After the derivation, we substitute x = 1 into (3 + 2A + 1) / 2 = 3 to get a = 1, and from the previous formula we get b = - 3

If limx →∞ [(2x ^ 2 + 3) / (x + 1) - ax + b] = 0, find the value of a and B

That formula = [2x ^ 2 + 3 + (x + 1) (- ax + b)] / (x + 1)
=[(2-a)x^2+(b-a)x+(b+3)]/(x+1)
Divide up and down by X
=[(2-a)x+(b-a)+(b+3)/x]/(1+1/x)
The denominator limit is 1
Then the molecular limit is 0
X tends to infinity
Then only 2-A = 0, B-A = 0
So a = b = 2

Given limx → 2 (x ^ 2 + ax + b) / (X-2) = 3, find the value of a and B
The steps in the book are as follows: 1. From the theme, we get limx → 2 x ^ 2 + ax + B = 3,2, limx → 2 (2x + a) = 3,3, so a = - 1, B = - 2?

Did you learn the law of lobita?
The molecular derivation is 2x + a
The denominator derivative is 1
So LIM (2x + a) = 3

Given limx → 2 (x ^ 2 + ax + b) / (2-x) = 3, find the value of a and B

lim(x->2)( x^2+ax+b)/(2-x) (0/0)
=lim(x->2) -(2x+a)
= -4-a = 3
a= -7
lim(x->2)(x^2+ax+b)/(2-x) =3
lim(x->2)(x^2-7x+b)/(2-x) =3
lim(x->2)[ (-x+5) + (b-10)/(2-x) ] =3
b-10 =0
b= 10