how to apply "$ \varepsilon$- $\delta$" definition of limit I am self studying calculus, and i came across a University quiz that required me to:


  Use the $ \varepsilon$- $\delta$ definition of limits to verify that the $\lim\limits_{x\to\pi} 3 = 3$?


how to apply such thing
(this definition have not been mentioned in the textbook I am refereeing to (calculus life saver))
The "$\varepsilon$ - $\delta$" definition of a limit is the following:

We say that $L$ is the limit of a function $f$ at $c$ (i.e., we say that $\lim_{x \to c} f(x) = L$) if, for every $\varepsilon &gt; 0$ there exists some $\delta &gt; 0$ such that, whenever $|x-c| &lt; \delta$, we have $|f(x) - L| &lt; \varepsilon$.

As stated in the comments, having a constant function makes things pretty easy, at least in terms of figuring out what $\delta$ needs to be. Following is the required proof:

Let $\varepsilon &gt; 0$, and set $\delta = 1$. (In this specific case, I can choose any positive number I want for $\delta$, but I'm choosing 1 just to be concrete.) Our function $f$ is a constant function, meaning that for all $x$, we have $f(x) = 3$. Therefore, whenever $|x - \pi| &lt; 1$, we see that $|f(x) - 3| = |3 - 3| = 0 &lt; \varepsilon$ as required. 

If you have any questions about this proof, please ask. If it's your first time seeing an $\varepsilon$ - $\delta$ proof (and it sounds like it would be), I would be surprised if it wasn't a bit confusing.
For any $\epsilon &gt; 0$ let $\delta = 5.6784\times 10^{237}$.

If $|x - \pi| &lt; \delta$ then $|3-3| = 0 &lt; \epsilon$.

So $\lim_{x\to \pi} 3 = 3$

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This follows the definition precisely.

The definition is literally that 

$\lim_{x\to a} f(x) = L$ if for any $\epsilon &gt;0$ we can find a $\delta$ (based upon $\epsilon$) so that whenever $|x-a|&lt; \delta$ it will follow that that $|f(x) -L| &lt; \epsilon$.  

What this means is that if we want to get $f(x)$ as close as we want to $L$ ($|f(x) - L|&lt; \epsilon$) we can achieve that by take $x$ within a range of $a$ ($|x-a| &lt; \delta$).  In general, the closer we want to get $f(x)$ to $L$ we will usually need to take a narrower range. (If we cant find such a range, that would mean $f(x)$ doesn't get that close to $L$ after all.  If we can always find a range no matter how small $\delta$ is then that means $f(x)$ gets close to $L$ as $x$ gets close to $a$.

But if $f(x) =3$ is a constant function then $|f(x) - 3| = 0$ always and we always have $f(x)$ as close to $3$ as possible.  SO to find the range $\delta$ that assures this is trivial.  $\delta$ could be anything.  Whatever $x$ we choose $|f(x) -3| = 0 &lt; \epsilon$.

So I chose $\delta$ to be anything I wanted.  I like weird and ridiculous numbers, so I chose $5.6784\times 10^{237}$.