# Strange limit question

#### jdstokes

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be given. Let $L$ be a real number. State the condition for saying that as $x$ tends to $a$, the limit of $f(x)$ is not $L$. The statement ought to begin with "Given there exists $\epsilon > 0$".

Best guess: $\lim_{x \rightarrow a}f(x) \neq L$ means, given there exists $\epsilon > 0$ there exists $\delta > 0$ such that $|x-a|<\delta \Rightarrow |f(x) - L| > \epsilon$. I'm really not sure about this, however.

Related Introductory Physics Homework News on Phys.org

#### whozum

Nonremovable singularity?

Pardon me?

#### saltydog

Homework Helper
First, lets go over the rule for the limit:

Given $\epsilon>0$, there exists a $\delta>0$ such that for all x satisfying:

$$|x-a|<\delta$$

we have:

$$|f-L|<\epsilon$$

Now, what happens if L is not the limit?

Then there should exists an $\epsilon>0$ such that for some $\delta>0$ and all x satisfying:

$$|x-a|<\delta$$

we have:

$$|f-L|>\epsilon$$

Make sure you understand these two differences. If L is the limit, by choosing $\delta$ small enough I can make the difference between L and f as small as I want.

If L is not the limit, then no matter how small I make $\delta$, I can always find an $\epsilon$ such that the difference between f and L will be larger.

#### jdstokes

So I was correct? Great, thanks for your help saltydog.

"Strange limit question"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving