Try to put $h = 0$ into a difference quotient before simplifying and you get $\frac{0}{0}$, which means nothing. Yet the values clearly settle down as $h$ gets small. A limit describes that settling. Holes in rational functions are the simplest place to see it.
Key idea
$\displaystyle\lim_{x \to a} f(x) = L$ means: $f(x)$ gets as close to $L$ as we like when $x$ is close enough to $a$, with $x \ne a$. The limit is about the values near $a$. It does not care what $f(a)$ is, or whether $f(a)$ exists at all.
Why it works
Take $f(x) = \frac{x^2 - 1}{x - 1}$. At $x = 1$ it is undefined. For every other $x$, $f(x) = \frac{(x - 1)(x + 1)}{x - 1} = x + 1$, so its graph is the line $y = x + 1$ with a single hole at $(1, 2)$. Near $x = 1$ the values crowd around $2$: $f(0.99) = 1.99$, $f(1.01) = 2.01$. So the limit is $2$ even though $f(1)$ does not exist.
Worked examples
Worked example: A hole
Find $\displaystyle\lim_{x \to 2} \frac{x^2 + x - 6}{x - 2}$.
- Substituting gives $\frac{0}{0}$: a hole, not an answer.
- Factor: $\dfrac{(x - 2)(x + 3)}{x - 2} = x + 3$ for $x \ne 2$.
- As $x \to 2$, $x + 3 \to 5$.
Answer: $5$
Worked example: The value at the point does not matter
$f(x) = 2x$ for $x \ne 2$, and $f(2) = 7$. Find $\displaystyle\lim_{x \to 2} f(x)$ and $f(2)$.
- Near $2$ (but not at it), $f(x) = 2x$, which approaches $4$.
- The value at $2$ was set separately: $f(2) = 7$.
Answer: The limit is $4$; $f(2) = 7$
Worked example: No trouble at all
Find $\displaystyle\lim_{x \to 3} (x^2 - 2x + 5)$.
- A polynomial has no holes or jumps, so its values near $3$ approach its value at $3$.
- $9 - 6 + 5 = 8$.
Answer: $8$
Common mistake
Thinking a limit is just $f(a)$, and that it cannot exist where $f(a)$ is undefined. Both examples above break that idea: one limit exists at a hole, and the other differs from the function's value. Substituting works only for functions without breaks at $a$; when you get $\frac{0}{0}$, simplify and look again.