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

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

Limx ~ ∞ [ax ^ 3 + (B-1) x ^ 2 + 2 / x ^ 2 + 1] = limx ~ ∞ ax + B-1 = 1
∴a＝0 b＝2

Find limit limx → 1 (M / 1-x ^ M-N / 1-x ^ n)

If limx tends to positive infinity [(x ^ 2 + 3x + 4) / (x + 1) - ax + b] = 2, find a and B. can we use the law of lobita?
a=1,b=0

Find the general solutions of the following differential equations: (1) x & sup2; y ″ + XY ′ = 1; (2) y ″ = 3x power of E; (3) YY ″ - y ′ & sup2; = 0
Big one high number problem, do right add score, urgent~~~~~

1.x²y〃+xy′=1,xy''+1*y'=1/x ,(xy')'=（ln|x|)' ,xy'=ln|x|+c,y'=ln|x|/x+c/x y=0.5ln²|x|+cln|x|+c'
2. Y ″ = 3x power of E, y '= 3x power of e ^ 3 + C, y = 3x power of e ^ 9 + CX + C'
Note: (3x power of E) '= (3x power of E) * (3x)' = 3x power of 3E
So the original function of e to the power of 3x is e ^ to the power of 3x / 3 + C
3.yy〃-y′2=0 （ y'*y′ -yy〃）/y'2=0 (y/y')'=0 y/y'=c y'/y=c1 ( lny)'=c1 lny=c1x+c2
y=e^(c1x+c2)+c3

Y = x ^ 2 / (a ^ 2-B ^ 2x ^ 2) finding the n-th derivative of Y
I'll just calculate the first few first. Is there a simple way to deduce

Find the derivative of y = ln | 2x-3 |
Be more detailed. I can understand it

1. When 2x-3 > 0, i.e. x > 3 / 2, then y = ln (2x-3), let u = 2x-3
y'=(lnu)'(2x-3)‘=1/u*2=2/(2x-3)
2. When 2x-3

The derivative of y = ln (2x + 5)

y'=(2x+5)'/(2x+5)=2/(2x+5)
Let t = 2x + 5, then y = LNT
y'=dy/dx=(dy/dt)(dt/dx)=[d(lnt)/dt][d(2x+5)/dx]=(1/t)*2=2/(2x+5)

Let y = (e ^ x + 1) / (e ^ x-1), find the derivative of Y?

y=(e^x+1)/(e^x-1)
So y '= [(e ^ x-1) * (e ^ x + 1)' - (e ^ x-1) '* (e ^ x + 1)] / (e ^ x-1) ^ 2
=[(e^x-1)*e^x-e^x*(e^x+1)]/(e^x-1)^2
=(-2e^x)/(e^x-1)^2
Not bad

Find the first derivative of y = (e ^ x-e ^ - x) / (e ^ x + e ^ - x)
4／（(e^x+e^-x)）^2
Please write the process

y'=[(e^x-e^-x)'(e^x+e^-x)-(e^x-e^-x)(e^x+e^-x)']/(e^x+e^-x)^2=[(e^x+e^-x)(e^x+e^-x)-(e^x-e^-x)(e^x-e^-x)]/(e^x+e^-x)^2=[(e^x+e^-x)^2-(e^x+e^-x)^2]/(e^x+e^-x)^2=4/(e^x+e^-x)^2

Y = x + e ^ x to find the derivative of X
How to calculate it is 1 / (1 + e ^ x)?
I started to figure out that it's also 1 + e ^ x, but the answer is 1 / (1 + e ^ x)

y'=1+e^x
If the answer is wrong, 1 / (1 + e ^ x) can be calculated