LIM (x tends to negative infinity) ((x ^ 2 + 1) ^ 1 / 2 + x) how to do?

LIM (x tends to negative infinity) ((x ^ 2 + 1) ^ 1 / 2 + x) how to do?

Multiplication √ (X & # 178; + 1) - x
The molecule is the square difference
So the original formula = LIM (x → - ∞) (X & # 178; + 1-x & # 178;) / [√ (X & # 178; + 1) - x]
=lim(x→-∞)1/[√(x²+1)-x]
Denominator tends to + ∞
So the original formula = 0

The function f (x) defined on R satisfies: F (1) = 1, and for any x ∈ R, the solution set of F '(x) log2x / 2 is (log2x means the natural logarithm of X with 2 as the base),

I can only give you some ideas
Let g (x) = f (x) - X / 2
Because f '(x)

Enjoy doing sth

enjoy doing sht.

How many hectares is 18500 mu

1 hectare = 15 mu
18500 Mu = 1233.333333 ha

If we know that the K-L power + 3K = 5 of X is a linear equation of one variable with respect to x, then k =, the solution of the equation

Since it is a linear equation of one variable, then the exponent of X must be 1
That means k = 2, so x = - 1

English translation
_______ _______ _______ he live _______ the bus station

___ how____ __ far_____ ___ does____ he live __ from_____ the bus station

Try to explain: the square difference of two consecutive odd numbers is twice the sum of the two consecutive odd numbers

It is proved that: let two continuous odd numbers be 2N-1, 2n + 1 (n is an integer), (2n + 1) 2 - (2n-1) 2 = (2n + 1 + 2n-1) (2n + 1-2n + 1) = 2 [(2n + 1) + (2n-1)], that is, the square difference of two continuous odd numbers is twice the sum of the two continuous odd numbers

Y = UX dy / DX = u + X (DU / DX)

Because y = UX
So dy / Du = y '
y'=u+u'x
u'=du/dx
So dy / DX = u + (DU / DX) X

The quotient of 13.9 divided by 69.5 plus x, and the sum is 5.6?

13.9÷69.5+x=5.6
0.2+x=5.6
x=5.6-0.2
x=5.4

Given that the definition field of function f (x) = (AX ^ 2 + 2aX + 1) is r, the range of a is obtained,

Did you miss the logarithm?
True number is greater than 0
That is: ax & # 178; + 2aX + 1 > 0
(1) When a = 0, 1 > 0 meets the meaning of the question;
(2) When a ≠ 0, then: a > 0, △ = 4A & # 178; - 4A