I have a few questions about the limit, and I am not afraid of getting tired of the master of high m

lim(x→2-) x-2|/[x-2]

lim(x→2-) x-2)/[x-2]

lim(x→2+) x-2|/[x-2]

lim(x→2+) x-2)/[x-2]

So the limit does not exist.

lim( x->2) [x^2+ax+b]/[x^2-x-2] (0/0)

lim( x->2) [2x+a] [2x-1] so a=2lim( x->2) [x 2+2x+b] [x 2-x-2] (0 0).

So when x=2, x2+2x+b=0

Substituting x=2 gives b=-8

3] Is x [x 3+1] an infinite amount that cannot be judged, so x is not known ?

e [-1 x] x->0 - is not an infinite mass. Yes.

The first limit is calculated by approximation, and the result is either 1, -1, or non-existent, depending on the definition.

The second question is not clear, is x trending towards 2? Or infinity? 0？

The first half of the third question is unclear, how much does x trend? 0？Infinity?

The second half of the question e [-1 x] x->0-, [1 x] tends to infinity, and e [-1 x] is infinity.

Visually inspect the landlord has not yet learned the law of Lopida, so the following method is introduced.

The first question uses the equivalent infinitesimal and the second question uses rationalization.

Entire? That's a bit of a lot, and it's all simple, but it's too time-consuming

Do you use tanx x, do the multiplication or division part in the whole equation can be proposed to be replaced, this denominator tanx is in the subtraction position, I don't know if you can understand this.

lim(sinx-x)/x³

lim(cosx-1)/3x²

lim-sinx/6x

Equivalent substitution should be replaced as a whole, and partial substitution is not allowed.

The formula for the second important limit, take a good look. The first one does not meet.

<> is like silver, judging limbs, and trying to rush to the world.

The right limit of f(x) at the difference 0 is -1 2, the left limit is 1 2, and the left and right limits are not equal, so there is no limit at 0 and 0 is broken.

When there is no cannacle, the highest power of the numerator is 3 times, the highest power of the denominator is 5 times, divided by the 3 times of the numerator, the denominator is 2 times, which is equivalent to 1 x 2, and when x tends to infinity, the limit of 1 x 2 is 0

1^2+2^2+3^2+……n^2=n(n+1)(2n+1)/6.

Proof: Consists of two numbers of cubes and formulas.

n+1)^3-n^3=3n^2+3n+1

n 3-(n-1) 3=3(n-1) 2+3(n-1)+1n-1) 3-(n-2) 3=3(n-1) 2+3(n-2)+1 are added to the two sides of the above equations.

n+1)^3-1^3=3(1^2+2^2+3^2+……n^2)3(1+2+3+……Hail sells +n).

1+1+1+…Pulse....+1)

3(1^2+2^2+3^2+…Bu Xiang....+n^2)=(n+1)^3-1-3n(n+1)/2-n

n+1)(n^2+2n-3n/2-n)

n+1)n(n+1/2)

n(n+1)(2n+1)/2.

1^2+2^2+3^2+……n^2=n(n+1)(2n+1)/6