A basic real analysis question on limit sup and limit inf Suppose $f$ be a function from $\Bbb R$ to $\Bbb R$. Now $\limsup_{t->x} f(t) - \liminf_{t->x} f(t) < \frac{1}{k}$. Does this imply that there is an open interval $I_x$ containing $x$ such that 
$$\sup\{f(t):t \in I_x\} - \inf\{f(t):t \in I_x\} < \frac{1}{k}$$. I think we can tell the following :
$$\sup\{f(t):t \in I_x(\epsilon)\} - \inf\{f(t):t \in I_x(\epsilon)\} < \frac{1}{k}+\epsilon$$ for any $\epsilon > 0$. But how the above ?

Yes, because you have strict inequality in your assumption. So denote $\displaystyle\limsup_{t \to x} f(t) - \liminf_{t \to x} f(t) = \alpha$. By assumption, $\alpha< \frac{1}{k}$, so we can take $\epsilon$ with $0 <\epsilon < \frac{1}{k}-\alpha$. Then it is true, as you think, that we can find an interval $I_x$ with $\sup\{f(t): t \in I_x\} - \inf\{f(t): t \in I_x\} < \alpha+\epsilon < \frac{1}{k}$.
