Q.In the following example, given $\epsilon>0$, find a $\delta>0$ such that whenever $|x-a|<\delta$, we must have $|f(x)-l|<\epsilon$: $\displaystyle\lim_{x\to 1}(x^2+x+1)=3$

Limit of a Function — From Intuition to Precision

Imagine you are walking toward a wall. You never actually touch the wall, but you can get arbitrarily close to it — a millimetre away, then a micrometre, then a nanometre. The wall is a limit: a value you approach but may never reach.

That is the core idea of a limit in mathematics. We ask: What value does a function $f(x)$ get close to, as $x$ gets close to some number $c$? The answer, if it exists, is called the limit of $f(x)$ as $x$ approaches $c$, written:

The function need not even be defined at $x = c$. The limit cares only about behaviour near $c$, not at $c$ itself.

Intuitive Example

Consider $f(x) = \frac{x^2 - 1}{x - 1}$. At $x = 1$, this function is undefined (division by zero). But what happens as $x$ gets very close to $1$?

The values crowd around $2$. So we say:

Even though $f(1)$ does not exist, the limit does.

One-Sided Limits

Approaching $c$ from the left (smaller values) and from the right (larger values) can give different results. We write:

• Left-hand limit: $\displaystyle \lim_{x \to c^-} f(x)$
• Right-hand limit: $\displaystyle \lim_{x \to c^+} f(x)$

If both one-sided limits exist and are equal, then the two-sided limit exists and equals that common value.

Example: $f(x) = \frac{|x|}{x}$ at $x = 0$.

• From the left ($x < 0$): $|x| = -x$, so $f(x) = -1$. Thus $\displaystyle \lim_{x \to 0^-} f(x) = -1$.
• From the right ($x > 0$): $|x| = x$, so $f(x) = 1$. Thus $\displaystyle \lim_{x \to 0^+} f(x) = 1$.

Since the one-sided limits differ, $\displaystyle \lim_{x \to 0} \frac{|x|}{x}$ does not exist.

The Precise Definition (Epsilon-Delta)

The intuitive idea — "gets arbitrarily close" — is made rigorous with the $\epsilon$-$\delta$ definition.