Limx → 0 = (tan3x) / X limit

Limx → 0 = (tan3x) / X limit

tan3x~3x,
Limx → 0 = (tan3x) / X limit = 3

limx->0 ln(1+3x)/sin4x

Substituting an equivalent infinitesimal for an equivalent infinitesimal
Original formula = Lim 3x / (4x)
=3/4

How to find the limit of TaNx / arcsinx when (x → 0)?

When x → 0, arcsinx ~ SiNx ~ x, TaNx = SiNx / cosx ~ x / cosx
Replace the original formula, equal to (x → 0) 1 / cosx = 1
That is to replace x ~ SiNx with the equivalent infinitesimal

S=1+2x+3x2+4x3+… +Nxn-1 (x ≠ 0 and X ≠ 1) = -___ ．

S=1+2x+3x2+4x3+… +nxn-1．xS=x+2x2+3x3+… +The difference between the two formulas is: (1-x) s = 1 + X + x2 + +Xn-1-nxn, ∵ x ≠ 1 and X ≠ 0, then s = 1-xn (1-x) 2-nxn1-x. so the answer is: 1-xn (1-x) 2-nxn1-x

Sum: 1 + X + 2x ^ 2 + 3x ^ 3 + +Nx ^ n (x is not equal to 0)

See clearly, the original problem is not to ask for x + 2x ^ 2 + 3x ^ 3 + +When x is not equal to 1, Sn = x + 2x ^ 2 + 3x ^ 3 + +nX^nXSn=X^2+2X^3+…… (n-1) x ^ n + NX ^ (n + 1) subtraction, (1-x) Sn = x + x ^ 2 + x ^ 3 + +X^n-nX^(n...

What is the derivative of y = (3x ^ 2 + 1) (4x ^ 2-3)?

f"(x)=(3x^2+1)"(4x^2-3)+(3x^2+1)(4x^2-3)"
=(6x)(4x^2-3)+(3x^2+1)(8x)
=24x^3-18x+24x^3+8x
=48x^3-10x

How to find the derivative of y = 3x-4x + 2,

y'=3*2x-4=6x-4

The second derivative of y = 3x ^ 4-4x ^ 2 + 5

A:
y=3x^4-4x^2+5
y'(x)=3*4x^3-4*2x=12x^3-8x
y''(x)=12*3x^2-8=36x-8
So:
The second derivative of y = 3x ^ 4-4x ^ 2 + 5 is y '' (x) = 36x-8

If f (x) is an odd function on R, and if x is greater than or equal to 0, f (x) = x (3x + 1), find the analytic expression of F (x) when x is less than 0

x0
So f (- x) = - x (- 3x + 1) = x (3x-1)
For odd functions, then f (x) = - f (- x)
therefore
x

It is known that y = FX is an odd function defined on R. when x is greater than 0, FX is equal to the square of x minus 3x plus 1

x> 0, f (x) = x ^ 2-3x + 1,
F (x) is an odd function, f (- x) = - f (x) = - x ^ 2 + 3x-1,
Let t = - x, then x = - t, that is, x0),
f(x)=-x^2-3x-1,(x