Q.In the following example, given $\epsilon>0$, find a $\delta>0$ such that whenever $|x-a|<\delta$, we must have $|f(x)-l|<\epsilon$: $\displaystyle\lim_{x\to 2}(2x+3)=7$

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.

In plain language: No matter how tight a tolerance $\epsilon$ you set around $L$, you can find a neighbourhood of width $\delta$ around $c$ (excluding $c$ itself) such that every $x$ in that neighbourhood gives an $f(x)$ within $\epsilon$ of $L$.

Example: Prove $\displaystyle \lim_{x \to 2} (3x - 1) = 5$.

We need: $|(3x - 1) - 5| < \epsilon$ whenever $0 < |x - 2| < \delta$.

Simplify: $|3x - 6| = 3|x - 2| < \epsilon$.

So $|x - 2| < \frac{\epsilon}{3}$. Choose $\delta = \frac{\epsilon}{3}$. Then:

If $0 < |x - 2| < \delta$, then $|(3x - 1) - 5| = 3|x - 2| < 3\delta = \epsilon$.

Done. The limit is proved.

Why Limits Matter

Limits are the foundation of calculus. Derivatives are limits of slopes of secant lines. Integrals are limits of sums of areas of rectangles. Without limits, calculus has no logical basis.

For now, remember: A limit answers the question "Where are you heading?" not "Where are you?"