High number problem, find the limit! Find the limit of TaNx / tan3x, where x approaches 90 degrees, using lobitar's law But I want to take steps

High number problem, find the limit! Find the limit of TaNx / tan3x, where x approaches 90 degrees, using lobitar's law But I want to take steps

My answer is similar to that of the second floor. I first change the whole formula into the form of SiNx and cosx, take the value of SiNx directly to get - cos3x / cosx, take the derivative by using lobida's law to get - 3sin3x / SiNx, and then substitute it into the value to get 3

Higher number limit problem If the real number B satisfies | B | > 1, LIM (1 + B + B ^ 2... + B ^ (n-1)) / b ^ n =? Sequence {an} 1 / N ^ 2 1 ≤ n ≤ 1000 n ^ 2 / (n ^ 2-2n) n > 1001 then {an} limit value A. 0 B, 1 C, 0or1 D, not present Thank you The second point is divided into braces in two cases

1. The formula of equal ratio summation for molecules: LIM (1 + B + B ^ 2... + B ^ (n-1)) / b ^ n n n → + ∞ = LIM (1-B ^ n) / [(1-B) · B ^ n] n → + ∞ = LIM (1 / b ^ n-1) (1-B) n → + ∞ |

Limit problem of high number problems Why is LIM (x → 1 -) e1 / X-1 = 0? (formula Description: when x is negative and tends to 1, X of e minus 1 / 1 power = 0)

X tends to 1-
So X-1 tends to zero-
So 1 / (x-1) tends to negative infinity
So the negative infinite power of e tends to zero
So limit = 0

Two high number problems for finding limits Question 1 lim2 ^ nsin (x / 2 ^ n) n approaches infinity (x is a constant that is not equal to zero) Question 2 limsin (x ^ n) / (SiNx) ^ n (Mn is a positive integer)

The answer to the first question is X. when n approaches infinity, sin (x / 2 ^ n) is equivalent to X / 2 ^ n, so it is X
Second, the inscription is not very clear and can't be done

A problem of finding the limit of high numbers LIM (n → infinity) n / (n ^ 2 + 3) + n / (n ^ 2 + 12) +... + n / (n ^ 2 + 3N ^ 2)= The answer is √ 3 · π / 9. Find the detailed steps

Do it with definite integral
Put n ^ 2 on the denominator, so
Original limit = Lim1 / N * ∑ 1 / [(1 + 3 (K / N) ^ 2]
=∫ [1 / (1 + 3x ^ 2)] DX integral interval o to 1
=1 / √ 3 arctan √ 3x| (o to 1)
=1/√3（π/3-0)
=√3·π/9

A high number problem about finding limit lim [1/（1-x）-3/(1-x) ³] x→1

First of all, your brackets are wrong. Change them, and then divide them all
1/（1-x）-3/(1-x ³）
=(x-1)(x+2)/[(1-x)(1+x+xx)]
=-(x+2)/(1+x+xx)
So when X - > 1, the limit is - 1