Limit versus function value

Worked example

Let $f(x)=\dfrac{x^2-1}{x-1}$ for $x\neq 1$, and $f(1)=4$. Find $\lim_{x\to 1}f(x)$ and $f(1)$. Explain why they need not match.

1. For $x\neq 1$, $f(x)=x+1$. That simplified expression matches the original wherever both are defined.
2. As $x$ approaches $1$, $x+1$ approaches $2$. Nearby points on the graph sit near height $2$.
3. The extra assignment $f(1)=4$ places a filled point at $(1,4)$. It does not change $f(1.001)$ or $f(0.999)$.
4. Therefore the limit is $2$ and the function value is $4$. Continuity will fail because these disagree.

$\lim_{x\to 1}f(x)=2$, while $f(1)=4$. The limit is not the filled point.

Contrasting non-example

A student says “the limit is $4$ because that is the filled-in point.” That confuses a single assigned output with nearby behaviour. Another non-example: concluding that the limit does not exist merely because $f(1)$ is undefined. Missing $f(a)$ is compatible with a perfectly ordinary two-sided limit.

Anannt Error Clinic

The mix-up we see most often, named kindly, with what to try instead.

Error Clinic — three frequent mix-ups.

1. Filled-point worship: treating $f(a)$ as the limit. The lab below is designed to break this. Move the filled point; ask whether nearby heights changed.
2. One-sided blindness: a jump discontinuity has two different side limits. Then the two-sided limit does not exist, even if both one-sided limits exist.
3. Table overconfidence: a table that never gets closer than $0.1$ can look like it approaches $2.1$ when the true nearby value is $2$. Tables estimate; they do not finish the argument unless the algebraic structure is used.

Anannt Method Choice

If the expression is a rational function that is undefined at $a$, try factoring before declaring that the limit does not exist. If a graph is given, read left-hand height, right-hand height, then the filled point, in that order. If a table is given, look at both sides and at whether the inputs are actually approaching $a$.