Given the function f (x) = x ^ 3, then when Lim △ → 0, (x + △ x) ^ 3-x ^ 3 / △ x What is (x + △ x) ^ 3-x ^ 3 / △ x equal to when seeking Lim △→ 0?

Given the function f (x) = x ^ 3, then when Lim △ → 0, (x + △ x) ^ 3-x ^ 3 / △ x What is (x + △ x) ^ 3-x ^ 3 / △ x equal to when seeking Lim △→ 0?

Definition of Derivative
The original formula = (x + △ x) ^ 3-x ^ 3 / △ x = (3x ^ 2 △ x + 3x △ x ^ 2 + △ x ^ 3) / △ x
=3x^2+3x△x+△x^2
When △→ 0, the latter two terms are 0
So it's 3x ^ 2

LIM (cosx-cos3x) / (5x) x tends to 0

Method 1: when the sum difference product cos α - cos β = - 2Sin [(α + β) / 2] · sin [(α - β) / 2] LIM (x → 0) (cosx-cos3x) / 5x = LIM (x → 0) 2sin2xsinx / 5xx → 0, SiNx can be replaced by X, the original formula = LIM (x → 0) 4x / 5 = 0

lim((x→0) (sinx/x)^1/(1-cosx)

The title should be x → 0 +
Using natural logarithm
lim((x→0＋) ln(sinx/x)^1/(1-cosx)
=LIM ((x → 0 +) ln (SiNx / x) / (1-cosx) (Equivalent Infinitesimal Substitution)
=LIM ((x → 0 +) 2 [lnsinx LNX] / x ^ 2 (Law of lobida)
=LIM ((x → 0 +) [cosx / sinx-1 / x] / X (general)
=LIM ((x → 0 +) [xcosx SiNx] / (x ^ 2sinx) (Equivalent Infinitesimal Substitution)
=LIM ((x → 0 +) [xcosx SiNx] / x ^ 3 (Law of lobida)
=lim((x→0＋) [ cosx-xsinx-cosx]/(3x^2)
=LIM ((x → 0 +)) [- xsinx] / (3x ^ 2) (Equivalent Infinitesimal Substitution)
=-1/3
therefore
lim((x→0＋) (sinx/x)^1/(1-cosx)
＝lim((x→0＋) e^[ln(sinx/x)^1/(1-cosx)]
=e^(-1/3)

lim(x->0)(cosx^2)^1/sinx^2
How to find the limit when x approaches 0?

The original formula = exp {LIM (x - > 0) ln (cosx ^ 2) / SiNx ^ 2} = exp {LIM (x - > 0) ln [1 + (cosx ^ 2-1)] / SiNx ^ 2} = exp {LIM (x - > 0) (cosx ^ 2-1) / x ^ 2} = exp {LIM (x - > 0) - (x ^ 4 / 2) / x ^ 2} = exp (0) = 1
Here we use the equivalent infinitesimal ln (1 + x) ~ x, 1-cosx ~ x ^ 2 / 2 (x - > 0)
If the questions are cos ^ 2x and sin ^ 2x, there is a simpler way
The original formula = LIM (x - > 0) (1-sin ^ 2x) ^ {(- 1 / sin ^ 2x) * - 1)} = e ^ {- 1} = 1 / E