高数求极限n趋于无穷大时,lim (1/n - sin(1/n))/ (1/n^2),lim (1/n - sin(1/
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高数求极限n趋于无穷大时,lim (1/n - sin(1/n))/ (1/n^2),lim (1/n - sin(1/n))/ (1/n^3)这一式子呢
求极限n趋于无穷大时,lim (1/n - sin(1/n))/ (1/n^2),
lim (1/n - sin(1/n))/ (1/n^3)这一式子呢?
求极限n趋于无穷大时,lim (1/n - sin(1/n))/ (1/n^2),
lim (1/n - sin(1/n))/ (1/n^3)这一式子呢?
n→∞,1/n→0+,所以可以令x=1/n→0+后,两极限是等价的(由海因定理保证)
lim (1/n - sin(1/n))/ (1/n^2)=lim (x - sinx)/ (x^2),和lim (1/n - sin(1/n))/ (1/n^3)=lim (x - sinx)/ (x^3),(x→0).
这样就可以通过罗比达法则,结合等价无穷小来求
最后求得:
lim (1/n - sin(1/n))/ (1/n^2)=lim (x - sinx)/ (x^2)=0
lim (1/n - sin(1/n))/ (1/n^3)=lim (x - sinx)/ (x^3)=1/6
lim (1/n - sin(1/n))/ (1/n^2)=lim (x - sinx)/ (x^2),和lim (1/n - sin(1/n))/ (1/n^3)=lim (x - sinx)/ (x^3),(x→0).
这样就可以通过罗比达法则,结合等价无穷小来求
最后求得:
lim (1/n - sin(1/n))/ (1/n^2)=lim (x - sinx)/ (x^2)=0
lim (1/n - sin(1/n))/ (1/n^3)=lim (x - sinx)/ (x^3)=1/6
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