What a limit is

Almost every idea in this course is a limit wearing a costume. Derivatives are limits of slopes, integrals are limits of sums, and series are limits of partial sums. Getting this chapter right pays for itself four times over.

Here is the whole idea in one sentence: a limit asks where a function is headed as the input approaches some point, and deliberately refuses to look at what happens when it arrives.

That refusal is the useful part, not a technicality.

The function that cannot answer for itself

Consider

$$f(x) = \frac{x^2 - 1}{x - 1}$$

At $x = 1$ this is $\tfrac{0}{0}$, which is not a number. The function has nothing to say there. But for every other $x$, the numerator factors and the fraction cancels:

So $f$ agrees with $x + 1$ everywhere except at $x = 1$, where $f$ is undefined and $x + 1$ is $2$. Approach $1$ from either side and $f$ gets arbitrarily close to $2$. It never is $2$.

We write

$$\lim_{x \to 1} f(x) = 2$$

and read it “the limit of $f(x)$ as $x$ approaches $1$ is $2$”. The statement is about the neighbourhood of $1$, not about $1$.

From both sides, or not at all

A limit exists only if the function approaches the same value from the left and from the right. Those two are the one-sided limits:

$$\lim_{x \to a^-} f(x) \quad \text{from below,} \qquad \lim_{x \to a^+} f(x) \quad \text{from above}$$

and the two-sided limit exists exactly when both exist and agree.

The standard counterexample is $\tfrac{|x|}{x}$. For positive $x$ it is $1$; for negative $x$ it is $-1$. It never settles.

The jump at zero is what “no limit” looks like. There is no single value the function is heading towards, so there is nothing for the limit to be — and saying “the limit is both $1$ and $-1$” is not an answer, because a limit is a single number or it does not exist.

Three ways a limit fails

Worth knowing by name, because a question that asks why a limit fails is asking which of these it is:

1. The one-sided limits disagree — a jump, like $\tfrac{|x|}{x}$ at $0$.
2. The function grows without bound — $\tfrac{1}{x^2}$ at $0$. It is heading somewhere, but $\infty$ is not a number, so the limit does not exist. Writing $\lim = \infty$ describes how it fails; it is not a value.
3. The function oscillates forever — $\sin\!\left(\tfrac{1}{x}\right)$ at $0$ crosses every value between $-1$ and $1$ infinitely often in any neighbourhood of zero, so it is never eventually close to anything.

The second and third of those are the shape you will meet most: a fraction that is $\tfrac{0}{0}$ at the point, rescued by algebra. The third needs multiplying by the conjugate rather than factoring, which is chapter 1.2’s business.

Reading a limit off a table

When the algebra is not obvious, a table narrows things down. For $\dfrac{\sqrt{1+x} - 1}{x}$ near zero:

Both sides converge on $0.5$, which is what the verified claim above says exactly — the limit is $\tfrac{1}{2}$.