# limit concept doubts

let $f(x)$ be a function and $a$ be a number. Suppose

$$\lim_{x \to a} f(x) = L$$

Then can I make the value of $f(x) = L$ by making $x$ sufficiently close to $a$? if it is possible...could you give me an example?

also why the $\epsilon$-$\delta$ definition is considered as simple and elegant?

## 2 Answers

It is not always the case that $f(a)=L$ just because $\lim_{x \to a} f(x) = L$. For example, the function $f(x) = \frac{\sin(x)}{x}$ is not valued at $0$, yet one of the standard proofs you will do in calculus (and as pointed out in Timbuc's answer) is to show that $$\lim_{x \to 0}\frac{\sin(x)}{x} = 1$$ On the other hand, the function $g(x) = x$ is such that $$\lim_{x \to 0} g(x) = g(0)$$ It all depends on how well the function is defined at the point where the limit is being taken. If a function is not behaved at all, the limit will not be defined. Then there is a middle ground where the function can be a bit crazy, but you can still determine a limiting value, even though the function isn't defined at that point. Then you can have "nice" functions who's limit at a point is the same as the function valued at that point.

$\varepsilon$-$\delta$ is "simple and elegant" because it concisely captures the key idea of limits. Namely, that if the limit exists, you can get as "close as you want." You can wiggle within $\delta$ away from a point, and you are guaranteed that your function is still within $\varepsilon$ of a point. Effectively, you end up trapping the limit in a box of area $4\delta \varepsilon$. Draw a picture of this to see what I mean. Since $\delta$ and $\varepsilon$ are arbitrary positive quantities, the "box" can be as small as you want, but never of area $0$. I think reading the logic $$0<|x-a|<\delta \implies |f(x)-f(a)|<\varepsilon$$ makes the definition of a limit look pretty simple. Further, when you solve for a value of $\delta$ that lets you prove the existence of a limit, I think that is quite satisfying! And the process seems elegant. For example, proving that $\ln(x)$ is continuous via $\varepsilon$-$\delta$ leads to a choice of $\delta$ that I find very satisfying.

• in this $\epsilon$ - $\delta$ defnition could this difference ( $f(x)$ - $f(a)$) become 0 at any time? – JINS K.JOHN Apr 1 '15 at 14:45
• @JINSK.JOHN No, they are strictly positive quantities. For example, $\lim_{n \to \infty}\frac{1}{n} = 0$ which means there exists $N\in \mathbb{N}$ such that $|1/n-0|<\varepsilon$ for all $n>N$, but it is never the case for any integer $n$ that $|1/n-0|<0$ – graydad Apr 1 '15 at 14:45

Not always. The limit can exist and equal $\;L\;$ without the function being equal to it. In fact, the limit can exist and the function may not even be defined on $\;a\;$ . For example,

$$\lim_{x\to 0}\frac{\sin x}x=1\;$$

yet the function isn't defined at $\;x=0\;$ .

• Even more! Limits are interesting when the function is not defined at $a$. :) – egreg Apr 1 '15 at 14:24
• but can i make the value of f(x) =L or in other words could epsilon become 0 in epsilon - delta defnition? – JINS K.JOHN Apr 1 '15 at 14:39
• No, $\;\epsilon>0\;$ always, it can't become zero – Timbuc Apr 1 '15 at 15:14