The limit of LIM (SiNx ^ m / (TaNx) ^ n) at x → 0 is solved by the property of equivalent infinitesimal

The limit of LIM (SiNx ^ m / (TaNx) ^ n) at x → 0 is solved by the property of equivalent infinitesimal

lim（x→0）(sinx^m/(tanx)^n)
=lim（x→0）x^m/x^n
=lim（x→0）x^(m-n)

Using equivalent infinitesimal to find the limit LIM (5x + (SiNx) ^ 2 - 2x ^ 3) / (TaNx + 4x ^ 2) When x tends to 0, neither of the above two floors will work I used to do it like this. After reading, I found that I could not do it like this, although the final answer was right

0

Find the limit LIM (x → 0) (TaNx SiNx) / sin? X

tanx-sinx=sinx/cosx-sinx
=[sinx(1-cosx)]/cosx
(tanx-sinx)/sin3x=(1-cosx/cosx)sin2x
=(1-cosx/cosx)/1-cos2x
=(1-cosx/cosx)/[(1-cosx)(1+cosx)]
=1/[cosx(1+cosx)]
When Lim tends to 0, it should be = 1 / 2

Find the limit Lim. [(TaNx SiNx) / (sin ^ 3x)] I can't get sleepy

Let's look at the first step: TaNx SiNx is the formula deformation, SiNx = TaNx * cosx, and then replace it with TaNx TaNx * cosx TaNx (1-cosx), then TaNx is equivalent to x, and 1-cosx is equivalent to
After 2x ^ 2, sin ^ 3x is equivalent to x ^ 3, when X - > 0, LIM (x - > 0) (x * 2x ^ 2) x ^ 3 = 1.2

Why Lim x → 0 (TaNx SiNx) / sin? X is equal to 1 / 2, including what formula is used

If it is not easy to write, the limit symbol will not be written
(tanx-sinx)/sin³x
=(1-cosx)/[cosxsin²x]
=(1-cosx)/[cosx(1-cos²x)]
=(1-cosx)/[cosx(1-cosx)(1+cosx)]
=1/[cosx(1+cosx)]
Let's take cosx = 1

lim(x→xπ)(tanx-sinx)/sin^3x The title is wrong x → 2 π

LIM (TaNx SiNx) / sin ^ 3x = limsinx (1 / cosx-1) / sin ^ 3x = LIM (1-cosx) / [sin ^ 2x * cosx] = lim2 (SiNx / 2) ^ 2 / [4 (SiNx / 2) ^ 2 ^ 2 * cosx] = Lim1 / [2 (cosx / 2) ^ 2 * cosx] when x → 2 π, cosx / 2 → - 1cosx → 1 original limit = 1 / 2

The limit of (E's X / 2) - 1 when Lim tends to infinity

=Two parts of LIM
=0

When Lim x tends to infinity, find the limit of (2x-1 / 2x + 1) to the power of X + 1

[limit symbol omitted]
The original formula = {[1 - 1 / (x + 1 / 2)] ^ - (x + 1 / 2)} ^ (- 1) * [1 - 1 / (x + 1 / 2)] ^ (1 / 2)
= e^(-1) * 1
= 1/e

When Lim x tends to infinity, find the limit of (2x-1 / 2x + 1) to the power of X

In this paper, we change the bracket into (2x-2 + 1 / 2x + 1) = (1 - (2 / 2x + 1)), (1 - (2 / 2x + 1)), X tends to infinity, which is the infinite power form of 1. According to the formula, only the limit of - 2 x / (2x + 1) is required, and the result is - 1, and the result is the - 1 power of E

Find the following limit. LIM (n tends to infinity) (2x power) * (sin * 1 / 2x power)

Put the "2x power" on the denominator, the limit becomes an important limit LIM (t → 0) Sint / T, so the limit is 1