Is the limit of a function just its value at that point?
Short answer: A limit asks what f approaches, not what f(1) is. Simplify to x + 1 → 2, even though f(1) is undefined.
The trap
The wrong idea
“The limit at a point is just the function's value there.”
It sounds reasonable, and a lot of JEE aspirants carry it into the exam. Here's a one-question check, then the explanation that makes it click.
Quick check
Try it before you peek
f(x) = (x² - 1)/(x - 1). As x → 1, the limit is:
- Undefined
- 2
- 0
Show the answer
B. 2
A limit asks what f approaches, not what f(1) is. Simplify to x + 1 → 2, even though f(1) is undefined.
The click
The limit of f(x) as x → a asks what value f(x) gets close to as x gets close to a, from both sides, without x ever being equal to a. The value f(a) itself plays no part. It can differ from the limit, or not exist at all.
Here f(x) = (x² − 1)/(x − 1). At x = 1 the expression gives 0/0, so f(1) is undefined. But for every x ≠ 1 we can factor and cancel: (x − 1)(x + 1)/(x − 1) = x + 1. Near x = 1, but not at it, f(x) behaves exactly like x + 1, which approaches 2. So the limit is 2. Numerically, f(0.99) = 1.99 and f(1.01) = 2.01.
The graph of f is the straight line y = x + 1 with a single hole at the point (1, 2). The limit is the height of the hole.
The limit and the value agree only when f is continuous at a; in fact that is the definition of continuity: lim f(x) as x → a equals f(a). For polynomials, sin x, eˣ and similar functions this holds wherever they are defined, which is why direct substitution usually works and why the misconception feels right. When substitution gives an indeterminate form such as 0/0, take it as a signal to simplify (factor, rationalise, use standard limits or L'Hôpital's rule), not as proof that the limit doesn't exist.
Got this one? The Knot Scan finds the misconception you don't know you have, in 8 questions.