Explanation:
For all x = 0,
2π
−1 < sin
0 < sin²
X
(²)
1 <1+ sin²
≤ 1, and
< 1, so
2π
X
Therefore, lim
For all x > 0, we have √ > 0, so we can multiply
the inequality by √ without changing the
inequalities.
√ã ≤ √ē (1 + sin²
Observe that lim
x → 0+
+0←x
≤ 2.
(²))
≤2√x.
√x = 0 and lim 2√√x = 0.
x →0+
2π
√ē (1 + sin² ( ²7 ) ) =
=0
X
