## Find the limit of Ln (1 + 2 ^ x) ln (1 + 1 / x) (x approaches positive infinity) Please explain in detail

Firstly, ln (1 + 1 / x) and 1 / X are equivalent to infinitesimals

So the original formula = ln (1 + 2 ^ x) / X

Rebida's Law: = (LN2) * 2 ^ X / (1 + 2 ^ x) = LN2

PS: the first floor is too wrong

### Let the constants a > 0 and b > 0, then ln (AX) / ln (BX) tends to infinity at the limit of X

∞ / ∞ type, the derivative of numerator and denominator can be obtained by Robida's rule, and the result is 1

### lim(2x-3)^20*(3x+2)^30/(5x+1)^50

2^20 * 3^30 /5^50.

### Why is the limit of n to the nth power when n is a positive integer and N approaches infinity 1? Please prove I'm a freshman. I just want to prove that its limit is one

For any Q > 1, N - > + ∞, N / (Q ^ n) = 0;

This means that when n - > + ∞, the exponential function grows faster than the primary function, which is a property often used. Typing is very troublesome. Can you find the proof of this by yourself? It should be easy to find it

Then it's simple

For any ε> 0,1+ ε> 1, so when n - > + ∞, N / ((1)+ ε）^ n) = 0; this means that when n is large enough, N0, then we can take ε= A / 2, the part I have proved to have n to the power of n

### It is proved that when n approaches infinity, the n-th power of (- 1) divided by the square of (n + 1) is equal to 0 This is his analysis. The absolute value of {(- 1) to the nth power divided by the square of (n + 1) minus 0} = 1 divided by the square of (n + 1), and then enlarge it, that is, it is less than 1 divided by (n + 1). I don't understand why to enlarge it. It seems that it's OK not to enlarge it

It's OK not to enlarge, but the expression for n is more complex

By 1 / (n + 1) ^ 2< ε Get n > ε^ (- 0.5) - 1, take the positive integer n > =[ ε^ (-0.5) -1]

If zoom in from 1 / (n + 1) ^ 2 < 1 / (n + 1) 1/ ε , Take positive integer n > = [1/ ε]

The latter method is much simpler to find n

### 0-th power limit of 0 Ask LIM (x ^ 2 + y ^ 2) ^ (x ^ 2 * y ^ 2) (x, y close to 0)

Let R2 = x ^ 2 + y ^ 2, the original formula = R ^ (2r2), let r = 1 / N, n approaches infinity, then the original formula = 1 / (n ^ (2 / N2)) = 1