What is the limit as x approaches 0 of #sin(1/x)#?

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What is the limit as x approaches 0 of #sin(1/x)#?

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Answer 1 · 2015-01-05T14:08:00

This limit does not exist, or with other words, it diverges.

You can see this by substituting #u=1/x#. Then, as #x# approaches zero, #u# approaches infinity.
Therefore:
#lim_(x->0)sin(1/x) = lim_(u->oo)sin(u)#
This limit does not exist, for the sine is a periodic fluctuating function.

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What is the limit as x approaches 0 of #sin(1/x)#?

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