limx→0 ln（1+3x）/sin2x

 limx→0 ln（1+3x）/sin2x

[ln(1+3x)]'/[sin2x]'=3/[2(1+3x)cos2x]→3/2,x→0

Find the limit limx → 0 (arcsinx) ^ 2 / X (√ 1 + x-1)

Limx tends to 01 / xsinx =?

1,

Limx (x tends to 0) xsinx =?

Limx (x tends to 0) xsinx = 0
provide free gifts
Limx (x tends to 0) x / SiNx = 1

Limx tends to 0, ln (1 + 2x) / xsinx

Using equivalent substitution: LIM (x → 0) ln (1 + 2x) / xsinx = LIM (x → 0) (2x) / (x · x)
=lim（x→0）2/x=∞

Using two important limits to find the limit of xsinx / 1 when limx → 0

Because x approaches 0, X is infinitesimal, and SiNx is bounded function, so infinitesimal times bounded function is infinitesimal

Find the limit of the following function: limx &; [1 / E - (x / x + 1) ^ x] x tends to 0

Because LIM (x - > 0) (x / x + 1) ^ x
=lim(x->0) [(1-1/(x+1))^(-(x+1))]^(-1)*(1-1/(x+1))^(-1)
=1/e
So the original formula = 0 * 0 = 0

High number problem "finding the derivative of function y = X4",

Is it y = x ^ 4? Because y = x ^ 4, y '= 4x ^ 3 can be directly substituted into (x ^ u)' = UX ^ (U-1). The derivative is to remember the formula: ① C '= 0 (C is a constant function); ② (x ^ U)' = UX ^ (U-1) (n ∈ q); memorize the derivative of 1 / X; ③ (SiNx) '= cosx; (cosx)' = - SiNx; (TaNx) '= 1 / (c)

How to find the derivative of this function

LGE is a constant and the derivative is 0
So ask for the front one
y'=x'*log2(x)+x*[log2(x)]
=log2(x)+x*(1/xln2)
=log2(x)+1/ln2

It's not urgent to find the derivative of the function y = e ^ (√ (2x + 1))

The derivative of a composite function is obtained slowly in order, y '= e ^ (√ (2x + 1)) * (1 / 2) (2x + 1) ^ (- 1 / 2) * 2
And simplify the equation
y'=e^（√（2x+1））*(2x+1)^(-1/2）