What is X - > infinite lime ^ x equal to?

What is X - > infinite lime ^ x equal to?

X approaches positive infinity and e ^ x is infinite
X tends to negative infinity, e ^ x limit is 0
So the lime ^ x limit does not exist when x approaches infinity

Ln (1 + ax & # 178;) / ax & # 178; limit
X tends to zero

X tends to zero
Then ax & # 178; tends to 0
So ln (1 + ax & # 178;) ~ ax & # 178;
So the original formula is 1

What is Lim [x - > 0 negative] ln (1 + ax ^ 3) / (x-arcsinx),

lim(x->0-) ln(1 + ax³)/(x - arcsinx)
= lim(x->0-) 3ax²/{[1-1/√(1-x²)](1+ax³)} 0-) 2x/[3ax²(1-1/√(1-x²)-x(ax³+1)/(1-x²)^(3/2)] 0-) 1/[3ax(1-1/√(1-x²)-(ax³+1)/(1-x²)^(3/2)]

Find the limit of Tan (2x + x ^ 2) / arcsinx when x tends to 0

Using Infinitesimal Substitution
Limtan (2x + x ^ 2) / arcsinx (x tends to 0)
=Lim2x + x ^ 2 / X (x tends to 0)
=LIM (2 + x) (x tends to 0)
=2

Finding the limit of X →∞ (SiNx + cosx)

SiNx + cosx = ノ 2 sin (PI / 4 + x) when x tends to infinity, there is no limit

The limit of (SiNx cosx) / e ^ x
X tends to be positive infinity
Is it possible to determine that the limit is 0

Yes
The numerator is a bounded quantity, the denominator is an infinite quantity, and the fraction is 0

It is clearer to solve the system of inequalities 1, - 3 (x + 1) - (x-3) < 8,2x out of 3 + 1-2 out of 1-x ≤ 1
2. - 2 ≤ 2x of 3 + 5 < 1 of 2
3、x+2＞0,x-3＞0,x-6≤0
4. The minimum integer solution of X ＞ - 2 / 3, x-4 ≤ 8-2x
All inequalities

1. - 3 (x + 1) - (x-3) < 83x + 3 + x-3 > 84x > 8x > 2x of 23 + 1-2 / 1-x ≤ 12x / 3 + 1-1 / 2-x ≤ 12x / 3-1 / 2-x ≤ 0-x / 3 ≤ 1 / 2x ≥ - 3 / 2  x > 22, - 2 ≤ 3 / 2x + 5 ＜ 1-2 ≤ 2x / 3 + 5 ＜ 1 / 2-7 ≤ 2x / 3 ＜ - 9 / 2-21 ≤ 2x ＜ - 27 / 2-21 / 2 ≤ x ＜ - 27 / 43, x + 2 ＞

Solving inequality 2x-2, x + 8 is less than 0

(2x-2) / (x + 8) 10, but not in line with x + 80, 2x-2-8, to sum up - 8

Finding the system of inequalities by X + 1 of three ≥ X-1 of two [X-2] + 8 ＞ 2x

1： 1 / 3 (x + 1) > = X-1 / 2 (x-1)
2X+2>=6X-3x+3
-1>=x
2： 3 (X-2) + 8 > 2x
3X-6+8>2X
x>-2
To sum up - 2

We know the inequality (kx-k2-4) (x-4) about x > 0, where k ∈ R. (1) find the solution of the above inequality; (2) is there a real number k such that there are only a finite number of integers in the solution set a of the above inequality? If it exists, find the value of K which makes the number of integers in a minimum; if it does not exist, explain the reason

(1) Let the solution set of the original inequality be a, when k = 0, a = (- ∞, 4); (2 points) when k ＞ 0 and K ≠ 2, the original inequality is reduced to [x - (K + 4K)] (x + 4) ＞ 0, ∵ K + 4K ＞ 4, (4 points) ≠ a = (- ∞, 4) ∪ (K + 4K, + ∞); (5 points) when k = 2, a = (- ∞, 4) ∪ (4, + ∞); (without separate analysis of the case of k = 2, the original inequality is reduced to [x - (K + 4K)] (x-4) ＜ 0, a = (K + 4K, 4); (7 points) (2) it is known from (1): when k ≥ 0, the number of integers in a is infinite; (9 points) when k < 0, the number of integers in a is finite; (11 points) because K + 4K ≤ − 4, if and only if k = 4K, that is, when k = - 2 (k = 2 rounded off), the number of integers in a is the least. (14 points)