4.2 Algebra of Limits of Functions 197(b) Exercise 16. •In Example 4.1.12 we proved that x -+O lim sin (l) x does not exist. In the next
example, we co nsider a closely related example.
Example 4.2.21 Use the squeeze principle to prove that lim x sin (.!.) = 0.
x-+O X
Solution: We start with the inequality, J sin tJ :::; 1, Vt E R Then Vx -/:-0,
lsin (~)I:::; 1. Multiplying both sides by JxJ, we have
JxJ lsin (~)I < Jx J; i.e.,Ix sin(~) I < Jx J.
But lim x = 0, so by the second squeeze principle, lim xsin (l) = 0. D
x-+0 x-+O x' '
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' 'yFigure 4.5' '
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y = x sin (1/x) ' '
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' ', y=-x
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