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Calculus
Evaluate the Limit limit as x approaches infinity of (sin(x)^2)/(x^2)
lim
x
→
∞
sin
2
(
x
)
x
2
lim
x
→
∞
sin
2
(
x
)
x
2
Step 1
Since
0
x
2
≤
sin
2
(
x
)
x
2
≤
1
x
2
0
x
2
≤
sin
2
(
x
)
x
2
≤
1
x
2
and
lim
x
→
∞
0
x
2
=
lim
x
→
∞
1
x
2
=
0
lim
x
→
∞
0
x
2
=
lim
x
→
∞
1
x
2
=
0
, apply the squeeze theorem.
0
0
[
x
2
1
2
√
π
∫
x
d
x
]
⎡
⎢
⎣
x
2
1
2
√
π
∫
x
d
x
⎤
⎥
⎦
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