174
Chapter 2 | Limits
θ 1+cos θ
1−cos θ θ
1−cos θ θ
lim θ →0
= lim
· 1+cos
θ →0
1−cos 2 θ θ (1+cos θ ) sin 2 θ θ (1+cos θ ) θ · sin sin θ
= lim
θ →0
= lim
θ →0
θ 1+cos θ
= lim
θ →0
=1· 0 2 =0.
Therefore,
1−cos θ θ
(2.19)
lim θ →0
=0.
1−cos θ sin θ .
2.20
Evaluate lim θ →0
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