What is 1-cos2x / xsinx equal to when we find the limit limx 0

What is 1-cos2x / xsinx equal to when we find the limit limx 0

For reference~

Find the limit limx of the following function, which tends to π xsinx / (x - π)

lim(x->π) xsinx/(x-π） (0/0)
=lim(x->π) (xcosx + sinx)
=-π

Limx → infinity 3x / X & # 178; + 5 =?

By using the law of Robita, the derivation of the numerator denominator is obtained. When 3 / 2x x tends to infinity, the derivative tends to 0, and the original function tends to 0

The derivative of F (x) = x (x + 1) (x + 2)... (x + n) at x = 0?

f(x)=x(x+1)(x+2)...(x+n)
Is a polynomial of degree n + 1
therefore
F '(x) is a polynomial of degree n
F '(0) is the constant term of F' (x)
The coefficient of the linear term x of F (x) is
1*2*3*...*n=n!
[choose one X and N constants from N + 1 factors
Because x = x + 0
So we can only select X in X + 0 and the constant term in other n factors]
So the constant term of F '(x) is n!
f'(0)=n!

The derivative of X (x-1) (X-2). (x-99) f '(0) =?

What is the derivative of (a ^ X / LNA)?

=(a^x)'/lna
=(a^x*lna)/lna
=a^x

Derivative of a ^ X / LNA

Formula (a ^ x) '= a ^ x * LNA
So (a ^ X / LNA) '
=1/lna*(a^x)'
=1/lna*a^x*lna
=a^x

What is the derivative of a * LNA?

If a is a constant
The derivative of a * LNA is 0
If we take the derivative of a
(a*lna)'=a'lna+a*(lna)'=lna+a-1/a=lna+1

How to prove the derivative of Cotx with the definition of derivative?
I can't prove it by definition!

lim(△x→0)[cot(x+△x)-cotx]/[(x+△x)-x]
=lim(△x→0)[cos(x+l△x)sinx-sin(x+△x)cosx]/[(sin(x+△x)sinx)*△x ]
=lim(△x→0) -[sin[(x+△x)-x]/△x]*[1/[sin(x+△x)sinx]
=lim(△x→0) -(sin△x/△x)*[1/sin(x+△x)sinx]
lim(△x→0)sin△x/△x=1
=-1/(sinx)^2

To prove INX 〈 x 〈 x with E as the base (using derivative)

Let f (x) = x-lnx, so f '(x) = 1-1 / x, when 0e ^ 0-1 = 0, that is, G (x) in (0, positive infinity) is monotonically increasing, and G (0) = e ^ 0-0 = 1 > 0, so for all x, G (x) > 0, then G (x) > 0 is obtained