Finding the limit: LIM (x + e to x power) 2 / x power

Finding the limit: LIM (x + e to x power) 2 / x power

lim(x→0) (x+e^x)^(2/x)
= lim(x→0) [1+(x+e^x-1)]^(2/x)
= lim(x→0) [1+(x+e^x-1)]^[1/(x+e^x-1) * (x+e^x-1)*(2/x)] 0) (1+y)^1/y=e,y=x+e^x-1
= e^2lim(x→0) (x+e^x-1)/x
= e^2lim(x→0) [1 + (e^x-1)/x] 0) (e^x-1)/x = 1
= e^[2(1+1)]
= e⁴

Lim X-1 x + 1 / 2x2-x-1 Lim x-0 (x + e to the x power) to the 1 / x power Lim X-1 x to the n power - 1 / X to the m power - 1

Hello!
The beauty of Mathematics
lim(x→1) (x - 1) / (2x² - x - 1)
= lim(x→1) (x-1) / [ (x-1)(2x+1) ]
= lim(x→1) 1 / (2x+1)
= 1/3
lim(x→0) (x+e^x)^(1/x)
= lim(x→0) e^ [ ln(x+e^x) / x ]
= e^ lim(x→0) ln(x+e^x) / x
=E ^ LIM (x → 0) (1 + e ^ x) / (x + e ^ x) [Robida's Law]
= e²
lim(x→1) (x^n - 1) / ( x^m - 1 )
=LIM (x → 1) n x ^ (n-1) / m x ^ (m-1) [Robida's Law]
= n / m

The number a is 36. The least common multiple of a and B is 360, the greatest common divisor is 4, and what is B

40

7) If there is a definition statement: int K1 = 10, K2 = 20;, after executing the expression (K1 = K1 > K2) & & (K2 = K2 > K1), the values of K1 and K2 are () respectively
A) 0 and 1 b) 0 and 20 C) 10 and 1 D) 10 and 20

0 20
K1 > K2 does not hold, K1 = 0; the previous part of the operation is false, and the program directly executes the next statement, so K2 = K2 > K1 is not run, K2 is still equal to 20, so the output K1 = 0, K2 = 20

English book of Grade 7 Volume 1, page 36-53, translation in full? (FLTRP) translation in full, not just the text, all the English words

Fainted, foreign language teaching and Research Press published English books, but a large number of what kind of specific title of you,

It is known that the parabola y = 1 / 2x & # 178; - x = - 3 / 2 has two intersections A and B with the X axis, intersects with the Y axis at point C, and the point E is on the positive half axis of the Y axis,
It is known that the parabola y = 1 / 2x & # 178; - x-3 / 2 has two intersections A and B with the X axis, intersects with the Y axis at point C, and the point E is on the positive half axis of the Y axis, and the triangle with a, O and E as the vertex is similar to the triangle with B, O and C as the vertex

The solution of y = 1 / 2x & # 178; - x-3 / 2 is x = - 1 or 3. A coordinate (- 1,0), B coordinate (3,0)
When x = 0, y = - 3 / 2, C point coordinates (0, - 3 / 2)
If two triangles are similar, then AO / CO = EO / Bo or AO / Bo = EO / Co
1. AO / CO = EO / Bo, 1 / (3 / 2) = EO / 3, EO = 2, the coordinates of point e are (0,2)
2. AO / Bo = EO / Co, 1 / 3 = EO / (3 / 2). EO = 1 / 2, e point coordinate is (0,1 / 2)

Find the 100th power of - 12 + 36-108 +... + 4x3

I don't know how old you are this year. If you've been in high school and studied series, this is a textbook version of the formula for summation of equal ratio series. The common ratio is - 3, the first term is - 12, and the number of terms is 100. The formula for summation of equal ratio series: SN = A1 * (1-Q ^ n) / (1-Q) (q is not equal to 1) (SN means the sum of the first n terms, A1 means the first term, Q means the common ratio

On the mutual transformation of polar equations in polar coordinates
The polar coordinate (6, π / 6) radius of the center is 6
Polar equation of a circle

According to the conversion formula between the two: x = PCOS @, y = PSIN @, where p represents the polar diameter, @ represents the polar angle, so in the center coordinate of the circle, P = 6, @ = π / 6. Substituting it, we can get x = 3 √ 3, y = 3 in the rectangular coordinate system, so the standard equation of the circle in the rectangular coordinate system is (x-3 √ 3) ^ 2 + (Y-3) ^ 2 = 36, The general equation obtained by expansion is x ^ 2-6 √ 3 + y ^ 2-6y = 0, and because P = √ (x ^ 2 + y ^ 2), x = PCOS @, y = PSIN @, the above equation is converted to p ^ 2-6 √ 3pcos@-6psin@=0 ,p-6√ 3cos@-6sin@=0 Wait until the conclusion

Solve several factoring problems in mathematics
7(a+b)2-5(a+b)-2
X6-Y6-2X3+1
4XY+1-4X2-Y2
One more question. The N + 2 power of a plus the N + 1 power of a, B minus the n power of 6a, B2

7 (a + b) 2-5 (a + b) - 2 = (7a + 7b + 2) (a + B-1) 7 21 - 1x6-y6-2x3 + 1 = (x6-2x3 + 1) - y6 = (x ^ 3-1) ^ 2-y ^ 6 = (x ^ 3-1-y ^ 3) + (x ^ 3 + y ^ 3-1) 4xy + 1-4x2-y2 = 1 - (4x ^ 2-4xy + y ^ 2) = 1 - (2x-y) ^ 2 = (1 + 2x-y) (1-2x + y) = = n + 2 power of a plus N + 1 power of a B

Given the function y = Y1 + Y2, and Y1 = 2x + m, y2 = 1 / (m-1) * x = 3, the ordinate of the intersection of two function images is 4, (1) find the functional relation of Y with respect to X
(2) If the image of the function intersects two coordinate axes a and B, calculate the area of △ AOB

Y1 = 2x + m, y2 = 1 / (m-1) * x + 3, the intersection of the two images is 4
Substitute y2 = 1 / (m-1) * x + 3 to get x = 1 / (M + 1)
Substitute x = 1 / (M + 1) into Y1 = 2x + m to get
M = 2 or M = 3
So the functional relationship between Y and X is y = 2x + 1 / x + 5 or y = 2x + 1 / 2x + 6