When Lim n tends to infinity, 1 / (n + 1) + 1 / (n ^ 2 + 1) is open to the second power +... + 1 / (n ^ n + 1) to the nth power

When Lim n tends to infinity, 1 / (n + 1) + 1 / (n ^ 2 + 1) is open to the second power +... + 1 / (n ^ n + 1) to the nth power

Zero
When n tends to infinity, the smaller the whole fraction, the whole formula tends to 0

What is the limit to the nth power of LIM (1-N / 1)? Masters``

This limit is equal to 1 / E. e is the base of natural logarithm, e = 2.718281828459045 It is a very important constant in mathematics

LIM (n →∞) (n + 1) (n + 2) (n + 3) / 5n3 power + n limit?

lim(n→∞) (n+1)(n+2)(n+3)/(5n³+n)
=LIM (n →∞) (1 + 1 / N) (1 + 2 / N) (1 + 3 / N) / (5 + 1 / N 2). Divide the numerator and denominator by N 3 at the same time
=1/5

Finding the limit of sequence: the nth power of LIM (n - ∞). (1-1 / N)

This problem is dealt with by the deformation of the important limit
LIM (1-1 / N) ^ n = ((1 + 1 / (- n)) ^ - n) ^ - 1 from the deformation of important limit, LIM (1 - 1 / N) ^ (- n) = e can be obtained
So the original formula = e ^ - 1 = 1 / E

Limit: LIM (1 + 1 / 2 + 1 / 4 + +1 / 2 to the N-1 power)

Original formula = LIM (1-1 / 2 ^ n) / (1-1 / 2)
=lim(2-1/2^(n-1))
=2

The 1 / x ^ 2 power of LIM (x → 0) ((arctanx / x) =? In the limit of the definite formula, it is necessary to use the law of lobida visually

X → 0 LIM (arctanx / x) ^ (1 / x ^ 2) = Lim e ^ ln (arctanx / x) ^ (1 / x ^ 2) = e ^ Lim ln (arctanx / x) ^ (1 / x ^ 2) consider Lim ln (arctanx / x) ^ (1 / x ^ 2) = Lim ln (arctanx / x) / x ^ 2 = Lim ln (1 + arctanx / x-1) / x ^ 2 according to the equivalent infinitesimal: ln (1 + x) ~ x = LIM (AR

Limit calculation by lopeda's rule: Lim x → 0 a ^ X-B ^ X / X (a > 0, b > 0) As the title

lim x→0 a^x-b^x/x
=lim x→0 [a^x *lna - b^x *lnb] / 1
=lna-lnb
=ln (a/b)

Who knows what type of L? Obita's law is and explains the process in detail

0 times infinity?

When x tends to infinity, how much is the limit of X to the power of 1 / 2, how to find it? It is required to use the law of l'urbida to ask for the guidance of the great God! Hurry! X tends to be positive infinity

Let's go step by step. It's a little complicated. We need the limit in the title. We assume that the function in the title is f (x), because it's too troublesome to write!
Let f (x) find the logarithm, that is, LN [f (x)] = (LNX) / X. let's first find the limit of this. According to the law of l'urbida, its limit is equivalent to the limit of the derivative of the numerator and denominator!
LIM (LNX) / x = LIM (1 / x) / 1 = LIM (1 / x) obviously, when x tends to infinity, the limit is 0
In other words, LIM (LNX) / x = 0
See clearly, our result is the value after the logarithm of F (x) in the title. What number takes the logarithm to get 0? Of course, it is 1
So the answer is 1

Find the limit of LIM (x tends to 0) (1 / X-1 / (the x power of E-1))

Find the limit of LIM (x tends to 0) (1 / X-1 / (the x power of E-1))
The above formula can be changed into:
(e^x-1-x)/(xe^x-x)
It belongs to the type of 0 / 0, and continuously applies the law of Robita
e^x/2e^x+xe^x
When x tends to 0, the expression tends to 1 / 2