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Today's Calculation Of Integral
2011 Today's Calculation Of Integral
761
761
Part of
2011 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 761
Source: 2011/10/31 National Defence Medical College
11/2/2011
Find
lim
n
→
∞
1
n
(
4
n
)
!
(
3
n
)
!
n
.
\lim_{n\to\infty} \frac{1}{n}\sqrt[n]{\frac{(4n)!}{(3n)!}}.
lim
n
→
∞
n
1
n
(
3
n
)!
(
4
n
)!
.
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