Computational Limit 1 with equivalent infinitesimal quantities, when x approaches 0 cos2x cos3x 1 x

1、lim[x→0]

cos2x-cos3x)/[√(1+x²)-1]lim[x→0]

cos2x-1+1-cos3x)/[√(1+x²)-1]lim[x→0]

cos2x-1)/[√(1+x²)-1]

lim[x→0]

1-cos3x)/[√(1+x²)-1]

cos2x-1 is equivalent to -(1 2)(2x) = -2x, and 1-cos3x is equivalent to (1 2)(3x) = (9 2)x

1+x )-1=(1+x) 1 2)-1 is equivalent to (1 2)x, so the above formula is:

Original = lim[x 0].

2x²/[(1/2)x²]

lim[x→0]

9/2)x²/[(1/2)x²]

2. E x-1 is equivalent to x, sinx is equivalent to x, and 1-cosx is equivalent to (1 2)x

Original = lim[x 0].

x²/[(1/2)x²]=2

cos(1 x) is equivalent to (1 2) (1 x) primitive=lim[x].

x²(1/2)(1/x²)=1/2

4、lim[x→0-]

f(x)lim[x→0-]e^x

lim[x→0+]

f(x)lim[x→0+]1+x)

f(0)=4

Therefore, the function is discontinuous at x=0 and is discontinuous.

1.The sum and differential products are used first, i.e., COS

cos=-2sin[(2]·sin[(2]), which changes the molecule to 2sin(5x2)sin(x2).

Divide the equation by (5x 2) (x 2) and multiply by (5x 2) (x 2) to make up the important limit, so that the numerator becomes 5x 2 2;

Next, because the denominator (1+x 2)-1 is equivalent to x 2 2;

So when x approaches 0, the limit of the function is 5

Equivalent to x, sinx is equivalent to x, 1-cosx is equivalent to x2 2, so the limit is 2

3.Swap 1 x for t, and from x to infinity t to 0, then it is easy to get the limit of 1 2

4.Discontinuity, is the point where the discontinuity can be reached.

by triangulation and difference formulas.

cosxcos2xcos3x

1/2)(cosx+cos3x)xos3x

1/4)cos2x+(1/4)cos4x+1/4+(1/4)cos6x

The original limit is (x->0).

1-(1/4)cos2x-(1/4)cos4x-1/4-(1/4)cos6x)/(1-cosx)

x->0

1-cosx~(1/2)x^2

Above = (1-(1 4)cos2x-(1 4)cos4x-1 4-(1 4)cos6x) (1 2)x 2

0 0 type) Lobida 0 0

Original limit = ((1 2)sin2x+sin4x+(3 2)sin6x) x=1+4+9=14

1-cosxcos2xcos3x=1-cos3x+cos3x(1-cos2x)+cos2xcos3x(1-cosx)~(3x)^2/2+(2x)^2/2+x^2/2=7x^2

Equivalent infinitesimal).

1-cosx~x^2/2

Original = lim7 x 2 (x 2 2) = 14

With the formula of double angle:

cos2a=1-2sin²a

1-cos2a=2sin²a

So: 1-cosx=2sin (x 2) 2 (x 2) x 2

So: 1-cosxEquivalent infinitesimal).is x 2

Equivalent infinitesimal is a relation between infinitesimal and refers to: in the same independent variable.

If the limit of the ratio of two infinitesimals is 1, then the two infinitesimals are said to be equivalent. The infinitesimal equivalence relation depicts that two infinitesimal approaches zero with equal velocity.

Equivalent infinitesimal substitution is a common method for calculating the unshaped limit, which can simplify the problem of finding the limit.

Answer: With the formula of double angle:

cos2a=1-2sin²a

1-cos2a=2sin²a

So: 1-cosx=2sin (x 2) 2 (x 2) x 2So: the equivalent infinitesimal of 1-cosx is x 2

x→0，1-cosx~x^2/2Commonly used infinitesimal substitution formula:

When x 0.

sinx~x

tanx~x

arcsinx~x

arctanx~x

1-cosx~1/2x^2

a^x-1~xlna

e^x-1~x

ln(1+x)~x

1+bx)^a-1~abx

1+x)^1/n]-1~1/nx

loga(1+x)~x/lna

The basic methods for finding the limit are:

1. In the fraction, the numerator and denominator are divided by the highest order, and the infinity is calculated by infinitesimal as infinitesimal and the infinitesimal is directly substituted with 0;

2. When the infinity root formula subtracts the infinite root formula, the molecule is rationalized;

3. Apply Lopida's law, but the application condition of Lopida's law is to become infinitely larger than infinite, or infinitesimal than infinitesimal, and the numerator denominator must also be a continuous derivative function.

When x tends to 0, the equivalent infinitesimal of 1-cosx is x 2, which can be obtained by the half-angle formula or Taylor's formula.

is 1 2 x square.

You can remember this, and by the way, you can remember the Taylor style, which is more useful.

FYI.

If you look at Taylor's formula for cosx, you can see which infinitesimal is equivalent to 1-cosx. But it's infinitesimal and not the only one, so you can't say what it is, the key depends on which precision you cut it to.

According to the Taylor formula.

cosx = 1-(1/2)x^2 +o(x^2)x->0

1-cosx = (1 2) x 2 + o(x 2)1-cosx is equivalent to (1 2) x 2

The equivalent infinitesimal of 1-cosx is 1 2x 2lim sinx x=1; (x->0)1-cosx=2*(sin(x 2)) 2The following limits tend to zero lim (1-cosx) (1 2*x 2)= 4* lim (sin(x 2)) 2 x 2=lim (sin (x 2) (x 2)) 2=1 glad to answer for you, please ask if you don't understand! Satisfied, thank you! o(..

x 0 at 2x 0 x 4 0

1-cos2x 1/2 (2x)^2 = 2 x^2sinx^4 x^4

After an equivalent infinitesimal substitution, the original becomes x 4x 0 when x 2 0

1 - cos (2 x^2) 1/2 ( 2 x^2 )^2 = 2 x^4

After the second substitution of the equivalent infinitesimal the original becomes 2 x 4 x 4 = 2

x 0lim (1-cos4x) x*sinx) because, cos4x=1-2sin 2(2x)=lim 2sin 2(2x) x*sinx) up and down divided by x 2 at the same time

lim 2sin^2(2x)/x^2 / x*sinx)/x^2=lim 8sin^2(2x)/4x^2 / x*sinx)/x^2=lim 8*sin^2(2x)/(2x)^2 / sinx/x)=lim 8*sin^2(2x)/(2x)^2 / lim (sinx/x)

According to the limit of the important one: lim sinx x=1

If you don't understand, please ask.