Limits, Discontinuities, And The Gap

A function can misbehave at a single input and still have a predictable trend nearby. Take

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

At $x=1$ the denominator is $0$, so $f(1)$ is not defined. If you only look at the function value, you get stuck.

Now look at what happens when $x$ is near $1$. Factor the top because it is the difference of squares.

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

For $x\neq1$, you can cancel the common factor $x-1$.

$$\frac{(x-1)(x+1)}{x-1}=x+1.$$

So when $x$ is close to $1$, the outputs are close to $1+1=2$, even though the original formula refuses to give an output at $x=1$. The number we are predicting is the limit.

Definition. We write $\lim_{x\to a}f(x)=L$ when $f(x)$ can be made as close as we want to $L$ by taking $x$ sufficiently close to $a$ (without requiring $x=a$).

Limits answer a specific question. What value does $f(x)$ approach as $x$ approaches $a$? That question still makes sense even if $f(a)$ is missing or wrong. What matters is the behavior near $a$, not the single point.

Holes and jumps as different failures

Two graphs can both be discontinuous at $x=a$ but for different reasons.

A removable discontinuity looks like a hole. The left side of the graph and the right side head toward the same height, but the function is either not defined at $a$ or defined at the wrong height. In that case the limit exists, and you could fix continuity by redefining $f(a)$ to equal the limit.

A jump discontinuity is different. As you approach $a$ from the left, the graph heads toward one height. As you approach from the right, it heads toward another. No single number can match both approach values, so the two sided limit does not exist, and there is nothing to fix by changing just one point.

To practice reading approach values from a graph, focus on where the curve is heading, not on whether there is an open or closed dot at $x=a$. Identify the left approach value, the right approach value, and then decide whether they agree.

One-sided limits decide the two-sided limit

The reason jump discontinuities break limits is that a two sided limit is really a meeting of two separate predictions.

The left-hand limit $\lim_{x\to a^-}f(x)$ tracks values with $x<a$ sliding toward $a$.

The right-hand limit $\lim_{x\to a^+}f(x)$ tracks values with $x>a$ sliding toward $a$.

The two sided limit exists exactly when those two one sided limits exist and are equal.

Golden rule. $\lim_{x\to a}f(x)$ exists if and only if $\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)$.

There are also two common ways a limit can fail even without a clean jump. The function might grow without bound near $a$, or it might oscillate so it never settles near a single height. In both cases, the two sided limit does not exist as a real number.

Classify each graph by checking the left side and the right side separately, then deciding what that says about the two sided limit.