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Find the limit limnsin (2 π en!) (n - > ∞)
By e ^ x = 1 + X + x ^ 2 / 2+ x^3/3!+...+ x^n/n! Know:
e=1+1+1/2!+ 1/3!+…+ 1/n!
SO 2 π en! Is an integral multiple of 2 π
So the limit is 0
The N5 power is the limit of the en power
Original formula = LIM (n - > ∞) n ^ 5 / e ^ n
=lim(x->+∞)x^5/e^x
=lim(x->+∞)5x^4/e^x
=lim(x->+∞)20x^3/e^x
=lim(x->+∞)60^2/e^x
=lim(x->+∞)120x/e^x
=lim(x->+∞)120/e^x
=0
As shown in the figure, make the vertical line en of diagonal BD from the vertex C of rectangular ABCD, the vertical foot is n, and intersect with the bisector AE of angle bad at E. verify AC = CE
Proof: because the quadrilateral ABCD is rectangular, angle DAC = angle DBC, angle DAB = angle ABC = 90 degrees, angle ABC = 90 degrees, CN is perpendicular to BD and N, so triangle CDN is similar to triangle BDC, angle DCN = angle DBC, angle DAC = angle DCN, because angle DAB = 90 degrees, AE bisects angle DAB, so angle DAE = angle BAE = 1 / 2
It is proved that the limit of LIM (n / N2 + 1) + (n / N2 + 2) + (n / N2 + 3). + (n / N2 + n) when n tends to infinity is 1
Just use the pinch theorem:
Set the original limit as I:
lim(n/(n^2+1))*n
What is the maximum value of the product obtained by multiplying any three numbers in integers - 5, - 3, - 1,2,4,6? What is the minimum value of the sum of any two numbers? Such as title
The maximum value of the product is 90 and the minimum value of the product is - 8
4. What are the maximum and minimum values of the product obtained by multiplying any two numbers in integers - 5, - 3, - 1, 2, 4 and 6?
Max. 4 * 6 = 24
Minimum (- 5) * 6 = - 30
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