When x tends to infinity, SiNx / x =? Is there no limit? How to write it?

When x tends to infinity, SiNx / x =? Is there no limit? How to write it?

0, the denominator tends to infinity, the whole also tends to 0

According to the definition of infinity, why is the reciprocal of limx tending to 0 (x (1 / 2 of SiNx)) not equal to 0?

In this case, whenever the limit of 1 / SiNx --- > ∞ or - ∞ or 1 / SiNx does not exist on X --- > k π, but x --- > ∞, the result does not exist, not 0

What is x + SiNx / X equal to when limx approaches infinity

It's (x + SiNx) / x, = 1

Limx tends to infinity, the answer of cosx / ex + E-X

When x tends to infinity, cos x is bounded, e ^ x + e ^ (- x) tends to infinity (no matter x tends to positive infinity or negative infinity), so the quotient of the two tends to zero

Seeking limx tends to 0 (cos2x) ^ [x ^ (2 / 3)] limit seeking process

Find the limit [5x & # 178; - 2 (1-cos2x)] / (3x & # 179; + 4tan & # 178; x) when x tends to 0

1-cos2x = 2Sin & # 178; X, so the original = (5x & # 178; - 4sin & # 178; x) / (3x & # 179; + 4tan & # 178; x) is divided by X & # 178; = [5-4 (SiNx / x) & # 178;] / [3x + 4 (TaNx / x) & # 178;] x tends to 0, then the limits of SiNx / X and TaNx / X are all 1, so the limit = (5-4) / (0 + 4) = 1 / 4

Limx tends to be positive infinity [√ (x ^ 2 + 2x) -√ (x ^ 2-2x)]

Multiplication √ (x ^ 2 + 2x) + √ (x ^ 2-2x)
Divided into = x & # 178; + 2x-x & # 178; + 2x = 4x
Original formula = lim4x / [√ (x ^ 2 + 2x) + √ (x ^ 2-2x)]
Divide up and down by X
=lim4/[√(1+2/x)+√(1-2/x)]
=4/(1+1)
=2

Given the inclination angle α = 3 π / 4 of line L, then the slope k of line L=______ ?

π = 180 degrees
3 π / 4 is 135 degrees
So the slope = Tan 135
=-1

A problem about slope in high school mathematics
If the origin is on the bottom edge of the isosceles triangle, what is the slope of the line where the bottom edge lies?
In this question, I suppose the equation of the straight line where the bottom edge is located is y = KX. Using the angle formula, I finally calculate k = 3 or K = - 1 / 3. The answer is that k = - 1 / 3 should be rounded off because when the slope k = - 1 / 3, the origin is not on the bottom edge of the isosceles triangle surrounded by the three straight lines. Why? How can I judge that when k = - 1 / 3, the origin is not on the bottom edge?

When k = - 1 / 3
First, find out the intersection of y = - X / 3 and X + Y-2 = 0, which is (3, - 1)
Then find the intersection of y = - X / 3 and x-7y-4 = 0, which is (6 / 5, - 2 / 5)
There is no point (0,0) in the line segment connecting these two points, and (0,0) should be on its extension line
So we have to give up

How to find the slope of a straight line

Three methods: (when the slope exists)
1. Known inclination angle a, slope k = Tana
2. If two points (x1, Y1) (X2, Y2) are known, then the slope k = (y1-y2) / (x1-x2)
3. If the direction vector (a, b) of a straight line is known, then the slope k = B / A