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?
 A: 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. 
A: 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 $\;x=0\;$ .
