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第二次習題 Exercise 2.5-32
- 1. 學號:U10011024 姓名:王冠倫
Exercise 2.5-32
sin(x + 1 π) − 1
3
lim
x→π/6 x − 1π
6
1
令y = x − 6 π
sin(x + 1 π) − 1
3
1
sin(y + 6 π) − 1
lim = lim
x→π/6 x − 1π 6
y→0 y
1 1
= lim [sin(y + π) − 1] lim
y→0 6 y→0 y
因為
1 1
lim [sin(y + π) − 1] = − ,
y→0 6 2
1 1
且 lim = −∞, lim = +∞,
y→0− y y→0+ y
所以,
sin(x + 1 π) − 1
3
1
sin(x + 3 π) − 1
lim = ∞, lim = −∞
x→π/6− x − 1π
6 x→π/6 + x − 1π
6
sin(x + 1 π) − 1
3
故 lim 不存在。
x→π/6 x − 1π
6
1