Let $f:(a,b) \to \mathbb{R}$ and $p \in (a,b)$.

In proving the following implication, I am unsure about one step

$\displaystyle \lim_{x \to p+}f(x) = l \land \lim_{x \to p-}f(x) = l \implies \lim_{x \to p}f(x) = l$

Supposing $\displaystyle \lim_{x \to p+}f(x) = l $ and $\displaystyle \lim_{x \to p-}f(x) = l$ we have, given $\epsilon > 0$

$\exists \delta_1 > 0 : \forall x \in (a,b)$ with $0 < x-p < \delta_1, \lvert f(x) - l \rvert < \epsilon$

$\exists \delta_2 > 0 : \forall x \in (a,b)$ with $- \delta_2 < x-p < 0, \lvert f(x) - l \rvert < \epsilon$

Set $\delta = \min \{ \delta_1 , \delta_2 \}$

This step seems intuitively obvious, does it suffice to say that

$\forall x \in (a,b)$ with $0 < \lvert x - p \rvert < \delta, \lvert f(x) - l \rvert < \epsilon$

since $\delta \leq \delta_1, \delta_2$

Sorry if this is a trivial question.

  • 1
    $\begingroup$ It is okay. Also note that $\delta>0$ $\endgroup$
    – drhab
    Apr 14 '14 at 9:09

This is correct, your proof is not perfect, but enough clear and understandable. To be more precise, you could finish that step as follows (I exaggerated a bit to emphasize my the intention):

We wanted to prove that there exists a particular $\delta$ given any $\varepsilon > 0$:

$$\forall \varepsilon > 0.\ \exists \delta > 0.\ \forall x \in (a,b).\ 0 < |x - p| < \delta \implies |f(x) - l| < \varepsilon. \tag{$\spadesuit$}$$

Let $\delta = \min\{\delta_1,\delta_2\}$, where $\delta_1$ and $\delta_2$ come from assumption for this particular $\varepsilon$ (this dependency is important, because the deltas are not any particular constants, instead they depend on the epsilon given).

Now, $|x - p| \leq \delta$ implies $|x - p| \leq \delta_1$ and $|x - p| \leq \delta_2$, hence $|f(x) - l| < \varepsilon$. Therefore, the $\delta$ as defined above satisfies the conditions of $(\spadesuit)$, that is, we proved the existence by constructing an appropriate element.

To summarize:

  1. You have to note the dependency between $\delta$ and $\varepsilon$.
  2. When dealing with an existential quantifier, instead of informally setting the variable to something, prove its existence by constructing it, e.g. "it exists because that formula calculates a number which has all the necessary properties".

I hope this helps $\ddot\smile$


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