Find the limit x tends to 0 LIM (cosx) ^ 1 / (x ^ 2) halo

Find the limit x tends to 0 LIM (cosx) ^ 1 / (x ^ 2) halo

Using logarithmic properties
(cosx)^(1/x^2)=e^[ln(cosx)^(1/x^2)]
=e^(1/x^2 * lncosx)
=e^(lncosx/x^2)
As long as the limit of the exponential part can be obtained, there are two methods
1、 Equivalent infinitesimal ln (1 + x) ～ x, 1-cosx ~ x ^ 2 / 2
lim(lncosx/x^2)=lim ln[1+(cosx-1)]/x^2
=lim (cosx-1)/x^2
=lim (-x^2/2)/x^2
=-1/2
2、 Derivation of numerator denominator and formula Lim SiNx / x = 1
lim(lncosx/x^2)=lim (-sinx/cosx)/2x
=lim (-1/2cosx)
=-1/2
So the original formula = Lim e ^ (lncosx / x ^ 2)
=e^lim(lncosx/x^2)
=e^(-1/2)

Find the limit LIM (x → 3) cosx

Because cosx is a continuous function with no discontinuity, its limit value at a point is the function value of that point. Therefore, the limit of cosx is cos3

LIM (x tends to infinity) e ^ - x ^ 2 * cosx

|cosx|≤1
lim(x->∞) e^(-x^2) .cosx
=0

How to find the limit of LIM (x - > 0) [((e ^ x) cosx-1-x) / (x ^ 3)] by Maclaurin formula?

The molecule is written as (e ^ x) cosx - e ^ x + e ^ X - 1 - x, and the method is simple

Calculate the following limit Lim / x-0 e - x + e x - 2 / 1-cosx

lim(x→0) (e^-x+e^x -2)/(1-cosx)
(x→0) e^-x+e^x-2 →0 1-cosx →0
lim(x→0)(e^-x+e^x-2)/(1-cosx)=lim(x→0) (e^x-e^-x)/sinx
x→0 e^x-e^-x →0 sinx→0
lim(x→0)(e^x-e^-x)/sinx=lim(x→0) (e^x+e^-x)/cosx=2
lim(x→0)(e^-x+e^x-2)/(1-cosx)=2

The value range of function y = 9 to the power of X + 4 × 3 to the power of X-1 is online

Let the x power of 3 be t (t is greater than 0)
Then f (x) = t squared + 4t-1
Then draw the graph to get the minimum value when x is taken as 0
The minimum value is: (- 1, positive infinity)

Given - 1 ≤ x ≤ 2, find the value range of function f (x) = 3 + 2 × 3 to the power of X + 1 - 9

Let a = 3 ^ X
Then 1 / 3

The function y = (1 / 3) is x ^ (2) - 2x power, and the range is () A.(0,3] B.(-∞,3] C.[-3,3] D.[3,+∞)

Y = (1 / 3) x ^ (2) - 2x power collocation method, y = (1 / 3) of (x-1) square + 1 (x-1) square + 1 is located in the upper right of 1 / 3 is the power of the range of value is to find (x-1) square + 1 value range is good (x-1) square + 1 value range also drawing method know that is greater than or equal to the y of this function y

The range of 1 / (x + 3) power of function y = 1 / 2

1/(x+3)≠0
So y ≠ (1 / 2) ^ 0 = 1
And the exponential function is greater than 0
So the range is (0,1) ∪ (1, + ∞)

Find the function y = (1 / 3) x power minus 4x, X belongs to the range of [0,5]

Calculate (1 / 3) x power and - 4x range respectively
(- 20 + (1 / 3) to the 5th power, 1]