A question on the argument : $\delta f(p) \le C$

Question

I'm confusing for the following argument is true or not:

Let $f(p) = \frac{1}{2-p}$. Then for any $1<p<2$, there exists a constant $C>0$ independent of $p$ and $\delta$ such that $$\delta f(p) \le C$$ if $\delta>0$ is small enough.

Answer 1

since $C>0$ is independent of $p,δ$ we can move it outside of the quantifiers for $p$, so your statement is equivalent to the assertion

$$∃ C>0 : ∀ p∈ (1,2), ∃ δ =δ(p ) :\quad f( p) \leq \frac{C}{δ(p)} $$

and this is true, In fact, any $C>0$ works, since then we can choose $δ( p) \geq \frac{C}{f( p)}$.