prove that lim x tens to 0 sin1/x does not exist
Answers
Proof. Assume that ∃A∈R s.t. limx→0sin(1/x)=A. There are three cases.
Case 1: |A|>1
By the definition of a limit, ∀ϵ>0, ∃δ>0 s.t.
|sin(1/x)−A|<ϵ whenever |x|<δ.
Restrict ϵ s.t. ϵ<|A|−1. Noting that in the interval (−δ,δ), sin(1/x) is at most 1,
|sin(1/x)−A|<|1−A|=|A−1|<ϵ<|A|−1.
However, |A−1|≥|A|−1. Thus, |A|≯1.
Case 2: |A|=1
Once again, by the definition of a limit, ∀ϵ>0, ∃δ>0 s.t.
|sin(1/x)−A|<ϵ whenever |x|<δ.
Restrict ϵ s.t. ϵ<1. Noting that at some point in the interval (−δ,δ), sin(1/x)=−A,
|−A−A|=2|A|=2<ϵ<1,
which is clearly false. Thus, |A|≠1.
Case 3: |A|<1
Once again, by the definition of a limit, ∀ϵ>0, ∃δ>0 s.t.
|sin(1/x)−A|<ϵ whenever |x|<δ.
Restrict ϵ s.t. ϵ<1−|A|. Noting that |sin(1/x)|−|A|≤|sin(1/x)−A|,
|sin(1/x)|−|A|≤|sin(1/x)−A|<ϵ<1−|A|.
Thus, |sin(1/x)|<1. However, in the interval (−δ,δ),∃x s.t. |sin(1/x)|=1. Thus, |A|≮1.
As shown through these cases, assuming that A exists always results in a contradiction. Thus, limx→0sin(1/x) does not exist.