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Today's Calculation Of Integral
2005 Today's Calculation Of Integral
63
Today's calculation of Integral 63
Today's calculation of Integral 63
Source: 1997 Hiroshima University
July 5, 2005
calculus
integration
limit
trigonometry
calculus computations
Problem Statement
For a positive number
x
x
x
, let
f
(
x
)
=
lim
n
→
∞
∑
k
=
1
n
∣
cos
(
2
k
+
1
2
n
x
)
−
cos
(
2
k
−
1
2
n
x
)
∣
f(x)=\lim_{n\to\infty} \sum_{k=1}^n \left|\cos \left(\frac{2k+1}{2n}x\right)-\cos \left(\frac{2k-1}{2n}x\right)\right|
f
(
x
)
=
lim
n
→
∞
∑
k
=
1
n
cos
(
2
n
2
k
+
1
x
)
−
cos
(
2
n
2
k
−
1
x
)
Evaluate
lim
x
→
∞
f
(
x
)
x
\lim_{x\rightarrow\infty}\frac{f(x)}{x}
x
→
∞
lim
x
f
(
x
)
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