How to find the limit of limx → infinity (5 + 1 / x-3 / x square) and lim → x → 1 (x cubic-3x + 2) / (x quartic-4x + 3) limit?

How to find the limit of limx → infinity (5 + 1 / x-3 / x square) and lim → x → 1 (x cubic-3x + 2) / (x quartic-4x + 3) limit?

The first one is that when x approaches infinity, the limits of 1 / X and 3 / x ^ 2 are zero, so according to the four operation rules of limit, the limit of the original formula is 5;
The second problem is 0 / 0 type inequality. According to lobida's rule, the upper and lower part of the semicolon are derived respectively, and the result is (3x ^ 2-3) / (4x ^ 3-4), which is still of type 0 / 0. If we apply the law of lobida again, we get 6 / 12x. Because x tends to be 1, the limit is 1 / 2

The limit of (x + 1) power of LIM (x →∞) [(2x + 3) / (2x-1)] The limit of (x + 1) power of LIM (x →∞) [(2x + 3) / (2x + 1)]

lim(x→∞）[（2x+3）/（2x-1）]^（x+1）
=lim(x→∞）[1+4/（2x-1）]^（x+1)
=lim(x→∞）[1+4/（2x-1）]^[（2x-1)/4]*[4*（x+1)/（2x-1)]
=lim(x→∞）e^[4*（x+1)/（2x-1)]
=e^2

What is the sum of the nth power of a and the nth power of B under the root sign of LIM? (0

It should be B ^ (n / 2)
lim√(a^n+b^n)=lim√{b^n[(a/b)^n+1]}
Because 0

Lim, n approaches the limit of infinity = (n + 2 under radical) - (n under radical)?

=Lim2/(√(n+2)+√n) (n->∽)=0

Find the limit of the nth power of [(a + B + C) / 3] when n tends to infinity The standard answer is ABC under the root three times. If you think your answer is correct,

Wrong number before
The limit of the nth power of [(a + N + C) / 3] tends to infinity at n is
ABC under three root signs
That is a ^ 1 / 3 * B ^ 1 / 3 * C ^ 1 / 3
a^(1/n)～1+(1/n)lna
a^(1/n)+b^(1/n)+c^(1/n)～3+(1/n)lnabc
=3(1+(1/3n)ln(abc))～3(abc)^(1/3n)
∴{(a^(1/n)+b^(1/n)+c^(1/n))/3}^n=(abc)^1/3

What is the limit of n-th root sign (1 + 2's nth power + 3's nth power) when n tends to infinity

Pinch method
3^n

Limit of high number: Lim n tends to infinity. Multiply N 2 to the nth power, and then divide it by the factorial of N. how can Lim n be infinite?

Using Stirling's formula, n! √ (2 π n) (n / E) ^ n
2^n/n-->∞
(2^n)^n/n!(2^ne/n)^n/√(2πn)-->∞

On the problem of higher numbers: is the result of LIM (n - > infinity) (a power of n) / logn related to a?

When a < 0, the molecular limit is 0, the denominator limit is infinite, and the limit is 0,
When a = 0, the molecular limit is 1, the denominator limit is infinite, and the limit is 0
a> When 0, the numerator denominator limit is infinite. Using the law of l'urbida, an ^ (A-1) / (1 / N) = an * n ^ (A-1) = a * n ^ A, infinity, limit does not exist

LIM (x tends to infinity) the difference between the nth power of 2 minus 1 divided by the nth power of 3 minus 1

To the nth power of the ratio of up and down to 2, the numerator approaches 1 and the denominator approaches infinity, so it finally approaches 0;

lim〔1+（1/2）+（1/4）+… [1 + (1 / 3) + (1 / 9) +... " 1 / (the nth power of 3)]

lim A= 1*(1-1/2^n)/(1-1/2)=2
lim B= 1*(1-1/3^n)/(1-1/3)=3/2
lim A*B= limA*limB=3