Why can't the common ratio and the first term of the equal ratio sequence be 0

Item n divided by common ratio equals item n-1. The common ratio cannot be 0 at the denominator position
In the same way
N + 1 divided by N is equal to common ratio Q, n (n = 1.2.3.4...) In the denominator position can not be 0

What is the root sign (x ^ 2 + A ^ 2) DX of indefinite integral?
RT
What is the indefinite integral 1 / root sign (x ^ 2 + A ^ 2) DX?

1.x/2√(x²+a²)+a²/2ln(x+√(x²+a²))+c
2.ln(x+√(x²+a²))+c

It is known that SiNx / X is a primitive function of F (x) to find ∫ x ^ 2F (x) DX

∫x^2f(x)dx
=∫x^2d(sinx/x)
=(x^2)(sinx/x)-∫(sinx/x)(2x)dx
=(x^2)(sinx/x)-∫2sinxdx
=(x^2)(sinx/x)+2cosx+C
C is an arbitrary constant

An original function of F (x) is e ^ - x, and the integral of XF '(2x) DX is obtained

f(x)=e^-x
f'(x)=-e^-x
f'(2x)=-e^-2x
∫xf'(2x)dx
=-∫xe^(-2x)dx
=1/2∫xe^(-2x)d(-2x)
=1/2*∫xde^(-2x)
=xe^(-2x)/2-1/2∫e^(-2x)dx
=xe^(-2x)/2+1/4∫e^(-2x)d(-2x)
=xe^(-2x)/2+1/4 *e^(-2x)+C
=e^(-2x)(x/2+1/4)+C

Let f (x) be a decreasing function in the domain (0, + ∞), and f (XY) = f (x) + F (y). F (1 / 3) = 1
(1) Finding the value of F (1)
(2) If f (x) + F (x + 2) < 2, find the value range of X

1. X = y = 1, xy = 1F (1) = f (1) + F (1) f (1) = 02, f (x) + F (x + 2) = f [x (x + 2)] = f (X & sup2; + 2x) 2 = f (1 / 3) + F (1 / 3 * 1 / 3) = f (1 / 9) f (X & sup2; + 2x) 1 / 99x & sup2; + 18x-1 > 0x (- 3 + √ 10) / 3 definition field x > 0, so x > 0, x + 2 > 0, so x > 0. To sum up, x > (- 3 + √ 10) / 3

If the function f (x) on the domain r satisfies that f (x + y) = f (x) + F (y) + 2XY (XY belongs to R) f (1) = 2, then f (- 2) and so on

x=y=0
f(0)=f(0)+f(0)+0=2f(0)
f(0)=0
x=1,y=-1
f(0)=f(1)+f(-1)-2=f(-1)
f(-1)=f(0)=0
f(-2)=f(-1-1)=f(-1)+f(-1)+2=2

Let f (x) be an increasing function on (0, + ∞), and f (XY) = f (x) - f (y)
(1) prove that f (x / y) = f (x) - f (y);
(2) if f (3) = 1 and f (a) > F (A-1) + 2, the value range of a is obtained
Correct the following question in the head is f (XY) = f (x) + F (y), not f (XY) = f (x) - f (y)

1、
f(xy)=f(x)+f(y)
Let x = y = 1, f (1) = 2F (1)
So, f (1) = 0
Let x = 1 / y, then f (1) = f (1 / y) + F (y)
That is: F (1 / y) + F (y) = 0
Then: F (1 / y) = - f (y)
So, f (x / y) = f (x) + F (1 / y) = f (x) - f (y)
The proof is complete
2、
f(xy)=f(x)+f(y)
Let x = y = 3, f (9) = 2F (3) = 2
Therefore, the inequality is: F (a) > F (A-1) + F (9)
That is: F (a) > F (9a-9)
Because the domain of definition is x > 0, so: a > 0,9a-9 > 0, get: a > 1;
Because f (x) is increasing, a > 9a-9
A

Let f (x) be differentiable, then Lim △ x → 0 [f ^ 2 (x + △ x) - f ^ 2 (x)] / △ x = --, the answer is 2F (x) f '(x), which has been solved analytically or calculated

lim [f^2（x+h）-f^2(x)]/h ( h→0)
=Lim [2F (x + H) f '(x + H) (H → 0) (0 / 0 Robida's law)
=2f(x)f'(x),
In the last step, f '(x) should be continuous
The correct solution is as follows
lim [f^2（x+h）-f^2(x)]/h ( h→0)
=Lim [f (x + H) + F (x)] [f (x + H) - f (x)] / h is the definition of derivative
=2F (x) f '(x), (f derivable must be continuous)

Let F X be continuous at x = 2 and lim (x tends to 2) (f (x) / (X-2)) = - 3, then
A f (x) is not differentiable at x = 2
B. not necessarily derivable
C. derivable but f ′ (2) ≠ - 3
D. derivable and f ′ (2) = - 3

LIM (X -- > 2) f (x) = 0 = f (2) (denominator -- > 0, the molecule must tend to 0, otherwise the limit does not exist)
Then f '(2) = LIM (X -- > 2) f (x) - f (2) / X-2 = LIM (X -- > 2) f (x) / X-2 = - 3

Why can't the common ratio and the first term of the equal ratio sequence be 0