Use squeeze theorem to find the limit as x approaches 0 of sin(1/x) and lim x->0 x sin (1/x)
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Use squeeze theorem to find the limit as x approaches 0 of sin(1/x) and lim x->0 x sin (1/x)

[From: ] [author: ] [Date: 11-09-30] [Hit: ]
However,Since |sin(1/x)| ≤ 1 for all nonzero x,|x sin(1/x)| ≤ |x| -|x| ≤ x sin(1/x) ≤ |x| for all nonzero x.Since lim(x→0) ±|x| = 0, the Squeeze Theorem implies that lim(x→0) x sin(1/x) = 0.I hope this helps!......
lim(x→0) sin(1/x) does not exist, because x→0, sin(1/x) oscillates more frequently between -1 and 1 inclusive (so there is no unique limit).

We can show this sequentially:
Pick two sequences {1/(πn)} and {1/(π/2 + 2πn)}, both of which converge to 0.

On one hand, sin(πn) = 0; so sin(1/x) on this sequence converges to 0, but
on the other hand, sin(π/2 + 2πn) = 1; so sin(1/x) converges to 1.
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However, lim(x→0) x sin(1/x) = 0 by the Squeeze Law:

Since |sin(1/x)| ≤ 1 for all nonzero x, we have
|x sin(1/x)| ≤ |x| <==> -|x| ≤ x sin(1/x) ≤ |x| for all nonzero x.

Since lim(x→0) ±|x| = 0, the Squeeze Theorem implies that lim(x→0) x sin(1/x) = 0.

I hope this helps!
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