The limit limx of higher numbers tends to the 1 / x-3 power of 3 (1-x / 3) Limx tends to the 1 / x-3 power of 3 (x / 3) Limx tends to the x power of infinity (X-2 / x + 2)

Look at the following answer,

In doubt,
Professional answers are not easy, please understand

Find the limit of the X * y power of (x * x + y * y) when x and Y tend to 0 Complete and detailed process derivation

First calculate the limit of (x ^ 2 + y ^ 2) ^|xy|
Because 01
Because (x ^ 2 + y ^ 2) ^ (XY) = (x ^ 2 + y ^ 2) ^ (±|xy|)
So (x ^ 2 + y ^ 2) ^ (XY) - > 1

Limit. N tends to infinity. What is the 1 / N power of N

y=N^(1/N)
lny=(1/n)*lnN=lnN/N
It is ∞ / ∞ type, using lobida's law
Molecular derivation = 1 / n
Denominator derivation = 1
When n tends to ∞, 1 / n tends to 0
So LNY limit = 0
So y limit = e ^ 0 = 1

A limit problem of higher numbers proves that the limit value of the 1 / N power of (1 + x) when x tends to zero is 1

Using a pinch theorem, when x > 0, it is between 1 and 1 + 1 / N * x; When x < 0, it is between 1 + 1 / N * x and 1. So the limit is 1
By definition, because | f (x) - a | ≤ 1 / N * | x |, it is determined by | f (x) - a ＜ ε Get | x | n ε, Just let the radius of the de centered neighborhood δ ≤n ε All right

X tends to the limit of a, (m power of X - m power of a) / (n power of X - n power of a)

Using Robida's Law:
=limmx^(m-1)/nx^(n-1)
=(m/n)a^(m-n)

Consult the limit of higher numbers to the x power of 1 When X - > infinity, what is the limit of the x power of 1, that is Lim 1 ^ x? Does this belong to the infinite form of Nobita's law about 1?

It's 1
However, if it is a number approaching 1 to the infinite power, it is not necessarily 1

The limit limx of higher numbers tends to the 1 / x-3 power of 3 (1-x / 3) Limx tends to the 1 / x-3 power of 3 (x / 3) Limx tends to the x power of infinity (X-2 / x + 2)