Is the following true? If 1f\frac{1}{f} is unbounded in every interval containing x=3x=3, then limx→ 3f=0\lim_{x\to\ 3}f = 0.

I know this is false since I have seen a counter example. However, I couldn’t figure out the flaw in my “proof”.

For any M>0M>0 and for any δ>0\delta>0, x∈(3−δ,3+δ)x\in (3-\delta,3+\delta) implies that |1f|>M\bigl| \frac{1}{f}\bigr|>M, which is equivalent to say that x∈(3−δ,3+δ)x\in (3-\delta,3+\delta) implies 1M>|f|\frac{1}{M}>|f|. Let ϵ=1M\epsilon= \frac{1}{M}. Thus, by taking very big MM and very small δ\delta, we can show that x∈(3−δ,3+δ)x\in(3-\delta,3+\delta) implies |f(x)−0|<ϵ|f(x)-0|<\epsilon ================= What is your counterexample? I think your problem is with the confusion of being "unbounded" and "having infinity as limit". You proof operates with the assumption that limx→31/f=∞\lim_{x \to 3} 1/f = \infty. – xyzzyz Oct 20 at 23:21 2 Thta's ∃x∈(3−δ,3+δ)\exists x\in (3-\delta,3+\delta). It's not for all xx in that interval, which would be required to show f→0f \to 0. – dxiv Oct 20 at 23:21 You could have something like sin(1x−3)(x−3)2\frac{\sin\left(\frac{1}{x-3}\right)}{(x-3)^2} that both grows and oscillates. – Joe Johnson 126 Oct 20 at 23:26 Simply, the unbounded-ness of 1f\frac{1}{f} is a neighbourhood of x=3x=3 does not imply that limx→3f(x)\lim_{x\to 3}f(x) exists. – Jack D'Aurizio Oct 21 at 1:34 Take f(x)=x−3f(x)=x-3 if xx is an irrational number, and f(x)=1f(x)=1 if xx is a rational number, for instance. – Jack D'Aurizio Oct 21 at 1:35 ================= =================