请大神求这道高数题的极限
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请大神求这道高数题的极限
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![请大神求这道高数题的极限](/uploads/image/z/115157-29-7.jpg?t=%E8%AF%B7%E5%A4%A7%E7%A5%9E%E6%B1%82%E8%BF%99%E9%81%93%E9%AB%98%E6%95%B0%E9%A2%98%E7%9A%84%E6%9E%81%E9%99%90)
lim cos(π/n)*lim∑i/(n^2 + i)
=1*lim∑i/(n^2 +i)
=lim∑i/(n^2 + i)
因为:
i/(n^2+n) ≤ i/(n^2 + i) ≤ i/(n^2 +1) 注:1≤ i ≤ n
所以,
∑i/(n^2 +n) ≤∑i/(n^2 + i) ≤∑i/(n^2 +1)
(1+2+┈+n)/(n^2 +n) ≤ ∑i/(n^2+i) ≤ (1+2 +┈ + n)/(n^2 + 1)
n(n+1)/[2*n(n+1)] ≤ ∑i/(n^2+i) ≤ n(n+1)/[2*(n^2 +1)]
1/2≤ ∑i/(n^2+i) ≤ (1 + 1/n)/[2*(1+1/n^2)] 注:分子、分母同除以 n^2
则:
1/2 ≤ lim∑i/(n^2+i) ≤ lim (1+0)/[2*(1+0)] = 1/2
所以,lim∑i/(n^2+i) = 1/2
因此,这道题的极限也等于 1/2.
=1*lim∑i/(n^2 +i)
=lim∑i/(n^2 + i)
因为:
i/(n^2+n) ≤ i/(n^2 + i) ≤ i/(n^2 +1) 注:1≤ i ≤ n
所以,
∑i/(n^2 +n) ≤∑i/(n^2 + i) ≤∑i/(n^2 +1)
(1+2+┈+n)/(n^2 +n) ≤ ∑i/(n^2+i) ≤ (1+2 +┈ + n)/(n^2 + 1)
n(n+1)/[2*n(n+1)] ≤ ∑i/(n^2+i) ≤ n(n+1)/[2*(n^2 +1)]
1/2≤ ∑i/(n^2+i) ≤ (1 + 1/n)/[2*(1+1/n^2)] 注:分子、分母同除以 n^2
则:
1/2 ≤ lim∑i/(n^2+i) ≤ lim (1+0)/[2*(1+0)] = 1/2
所以,lim∑i/(n^2+i) = 1/2
因此,这道题的极限也等于 1/2.