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!
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.
---------------
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!