Lim x approaches infinity for x^2sin[1/x]
I kept using L'hospital's rule and kept getting 0/0. Supposedly the answer is eventually supposed to equal infinity but I'm not getting that answer
ind the limit infinity for a function?
2016-04-19 2:48 pm
回答 (2)
2016-04-19 3:01 pm
✔ 最佳答案
x^2 sin(1/x) = sin(1/x) / (1/x^2)lim x-->∞ sin(1/x) /(1/x^2) = 0/0
=lim x-->∞ cos(1/x) (-1/x^2) / (-2/x^3)
= lim x-->∞ x cos(1/x) / 2
as x approaches ∞inity, cos(1/x) approaches cos(0) = 1
lim x-->∞ x (1/2)
= ∞
2016-04-19 3:03 pm
If 0 < u < π /2, then sin u > u/2.
Hence, if x > 2/π then sin (1/x) > (1/2)(1/x) = 1 / (2x)
So if x → ∞, eventually x > 2/π.
Then x^2 sin (1/x) > x^2 / (2x) = x/2, → ∞.
Hence, if x > 2/π then sin (1/x) > (1/2)(1/x) = 1 / (2x)
So if x → ∞, eventually x > 2/π.
Then x^2 sin (1/x) > x^2 / (2x) = x/2, → ∞.
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原文連結 [永久失效]:
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