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Limit of sin(1/x) as x approaches 0

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Evaluate the Limit limit as x approaches 0 of sin(1/x)

Taking the reciprocal: cos x ≤ sin x x ≤ 1. Since cos x, sin x x, 1 functions are even, then we conclude that: cos x ≤ sin x x ≤ 1, ∀ x ∈] − π 2, 0 [ ∪] 0, π 2 [. By using the Squeeze

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limit as x approaches 0 of x^4e^{sin(1/(x^2))}

x ⋅ sin ( 1 x) → 0. can be obtained in the same way by squeeze theorem. 0 ≤ | x ⋅ sin ( 1 / x) | = | x | ⋅ | sin ( 1 / x) | ≤ | x | → 0. once we know that, we can also proceed by standards
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