A problem of factorization Bisection of 5x (a-b) - 3Y (B-A)

A problem of factorization Bisection of 5x (a-b) - 3Y (B-A)

The square of 5x (a-b) - 3Y (B-A)
=5x(a-b)^2+3y(a-b)
=(a-b)(5ax-5bx+3y)

A = 3b-5, find 4 (3b-a) ^ 2-36b + 12a + 9

∵a=3b-5
∴3b-a=5
Then 4 (3b-a) & 178; - 36B + 12a + 9
=4(3b-a)²-12(3b-a)+9
=4*5²-12*5+9
=49

A ^ 4-A ^ 2B ^ 2 of a ^ 2-ab-2b ^ 2
Approximate points

What is the limit of x ^ 2 * sin (1 / x) at X - > 0?

0
X ^ 2 is infinitesimal
Sin (1 / x) bounded
So the limit is infinitesimal, and the limit is zero

The limit of X of (1 / x) sin (1 / x) to compact positive 0

The answer is positive infinity
1 / X tends to be positive infinity
And sin (1 / x) is a bounded variable
So the multiplication of two formulas is still positive infinity

The limit of sin √ x + 1-sin √ x, X tends to 0

Kid, you're here to make fun of me. I'll take 1 directly from the front and back row. If you want to deduce mathematically, you can bring x = 0 directly in. I guess you're hanging up miserably

Limx - > 1 sin (x-1) / X-1

1

When x tends to x 0, the right limit and the left limit of F (x) exist and are equal. What is the condition for the existence of the limit of F (x) tending to x 0
The book is filled with the necessary and sufficient conditions, but if x0 is the breakpoint that can be removed
(the accessible discontinuity is left limit = right limit, but not = the function value of the point, or there is no definition at the point.),

You don't fully understand the limit of function. When we discuss the limit of function f (x) when x → x0, we study the change of function value f (x) when x → x0 and X ≠ x0, which has nothing to do with whether f (x0) exists and how much f (x0) is equal to

If the function f (x) has a limit at point x0, then f (x) is continuous at point x0
A right B wrong

Wrong
...
For example, y = 0 (x ≠ 0)
lim x→0 y=0
But y is discontinuous at x = 0

Prove 1 / 2 + sum cosk (x-x0) = (sin (n) + 1 / 2) (x-x0) / 2 * sin ((x-x0) / 2)

The left side just needs to be multiplied by 2 * sin ((x-x0) / 2, and the sum difference formula of integration is used,
Left * 2 sin ((x-x0) / 2)
=sin ((x-x0)/2)-sin((x-x0)/2)+sin(3(x-x0)/2)-sin(3(x-x0)/2)+sin(3(x-x0)/2)
+...-sin((2n-1)*(x-x0)/2)+sin((2n+1)*(x-x0)/2)
=sin[(n+1/2)*(x-x0)]
The result is proved