# How $\delta_1$ and $\delta_2$ for two different limits at $a$ can be read as $\delta=\text{min}(\delta_1,\delta_2)$?

I am having trouble understanding a certain part of the proof on why a function cannot approach two different limits near $a$, so I will just list the relevant parts. If this is not enough/ambiguous then please tell me and I will type out the whole proof.

So, suppose we now have:

$$\text{if } 0<|x-a|<\delta_1, \text{ then } |f(x)-l|<\epsilon \hspace{5cm} (1)$$

and

$$\text{if } 0<|x-a|<\delta_2, \text{then} |f(x)-m|<\epsilon \hspace{5cm} (2)$$

and here's a quote from the text:

We have had to use two numbers, $\delta_1$ and $\delta_2$, since there is no guarantee that the $\delta$ which works in one definition will work in the other. But, in fact, it is now easy to conclude that for any $\epsilon>0$ there is some $\delta>0$ such that, for all $x$, $$\text{if } 0<|x-a| < \delta, \text{then } |f(x)-l| < \epsilon \text{ and } |f(x)-m| \lt \epsilon$$ we simply chose $\delta=\text{min}(\delta_1,\delta_2)$

I understand the need to use two distinct $\delta$. What I don't get is why selecting a $\delta$ that is the minimum of $\delta_1$ and $\delta_2$ will make that $\delta$ work in both (1) and (2). I mean, the limits are different so why would I expect that the delta that is the minimum of the two equations will satisfy both equations?

Thank you in advance for any help provided.

• Any $x$ such that $|x-a|\lt \delta$ automatically satisfies the condition $|x-a|\lt \delta_1$, also the condition $|x-a|\lt \delta_2$. If the distance of $x$ from $a$ is less than $\delta$, then it is less than $\delta_1$ and less than $\delta_2$. – André Nicolas Jul 24 '13 at 20:13

Your definition of limit is that for any $\epsilon>0$, there is some $\delta >0$ such that whenever $|x - a| < \delta$, $|f(x) - L| < \epsilon$. (This is exactly what you wrote.)
What you are overlooking is the less-than sign. By definition $\min(\delta_1, \delta_2) \leq \delta_1$ and $\min(\delta_1,\delta_2) \leq \delta_2$. Then by transitivity, whenever $|x - a| < \min(\delta_1,\delta_2)$, you know that $|x-a|$ is less than both $\delta_1$ and $\delta_2$; i.e., $|x-a|$ will work in both (1) and (2).
I hope this is not too wordy. It entirely boils down to $x < min(y, z) \implies x < y$ for any real $x,y,z$.
If $\delta=\min(\delta_1,\delta_2)$, then for $x$ s.t. $0 < |x-a| < \delta$, you have both $0<|x-a|< \delta \leq \delta_1$ and $0<|x-a|< \delta \leq \delta_2$; so $$|f(x)-l|<\epsilon \hspace{5cm} (1)$$ and $$|f(x)-m|<\epsilon \hspace{5cm} (2)$$