Most of math so far has asked: what is the value here? Calculus asks a sneakier, more powerful question: where is this function headed? A limit answers that — and the beautiful part is that a function can be heading somewhere it never actually reaches. Master this one idea and the rest of calculus opens up, because derivatives and integrals are both limits in disguise.
A function that points at an answer it can't reach
Look at this function:
\[ f(x) = \frac{x^2 - 1}{x - 1} \]Try to evaluate it at \( x = 1 \). The top becomes \( 1 - 1 = 0 \), the bottom becomes \( 1 - 1 = 0 \), and you get \( \tfrac{0}{0} \) — which is nonsense. The function is simply undefined at \( x = 1 \). There is a hole in the graph right where you most want an answer.
But watch what happens for every other value of \( x \). The top factors:
\[ \frac{x^2 - 1}{x - 1} = \frac{(x-1)(x+1)}{x - 1} = x + 1 \qquad (x \neq 1) \]So away from that single forbidden point, \( f \) is just the line \( x + 1 \). And a line doesn't care about holes — it heads somewhere definite. As \( x \) creeps toward 1, the output \( x + 1 \) creeps toward 2. The function never arrives at \( x = 1 \), but it is unmistakably headed for the value 2. That destination is the limit.
Watch it close in from both sides
You don't have to take this on faith — you can feel it numerically. Pick values of \( x \) sneaking up on 1 from the left (smaller) and from the right (larger), and read off \( f(x) = x + 1 \):
- \( x = 0.9 \;\Rightarrow\; f(x) = 1.9 \)
- \( x = 0.99 \;\Rightarrow\; f(x) = 1.99 \)
- \( x = 0.999 \;\Rightarrow\; f(x) = 1.999 \)
Climbing toward 2.
- \( x = 1.1 \;\Rightarrow\; f(x) = 2.1 \)
- \( x = 1.01 \;\Rightarrow\; f(x) = 2.01 \)
- \( x = 1.001 \;\Rightarrow\; f(x) = 2.001 \)
Settling down toward 2.
Both columns are squeezing in on the same number. We write this as:
\[ \lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2 \]Read it aloud: "the limit, as \( x \) approaches 1, of the function, is 2." Notice what the notation does not say. It never claims \( f(1) = 2 \) — in fact \( f(1) \) doesn't exist. The limit reports the destination, completely independent of whether the trip ever finishes.
To make that "approaching" motion concrete, drag the point below toward 1 from either side and watch how it can get arbitrarily close without ever having to land exactly on it.
In words A limit is the single value a function gets closer and closer to as the input gets closer and closer to some number — and it doesn't matter one bit what the function does, or fails to do, exactly at that number.
Two sides of the same approach
That table hinted at something important: \( x \) can sneak up on 1 from two different directions, and we tracked them separately. Each direction has its own name.
- The left-hand limit, written \( \displaystyle\lim_{x \to 1^-} f(x) \), is the value \( f \) heads for as \( x \) approaches from below (the little minus sign means "from the left").
- The right-hand limit, written \( \displaystyle\lim_{x \to 1^+} f(x) \), is the value \( f \) heads for as \( x \) approaches from above.
Here is the rule that ties them together, and it is the whole game:
\[ \lim_{x \to a} f(x) = L \quad\text{exists}\quad \Longleftrightarrow \quad \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L \]The two-sided limit exists only when both one-sided limits agree. If the function is heading for 2 from the left but for 5 from the right, there is no single destination — so the two-sided limit simply does not exist. The two scouts have to report back the same number.
Continuity: drawing without lifting your pen
Now we can pin down a word you've used loosely your whole life: a function being "smooth" or "unbroken." A function \( f \) is continuous at a point \( a \) when three conditions all hold at once:
- \( f(a) \) is defined — the function actually has a value at \( a \) (no hole).
- \( \displaystyle\lim_{x \to a} f(x) \) exists — the function is heading somewhere definite (both sides agree).
- They match: \( \displaystyle\lim_{x \to a} f(x) = f(a) \) — the place it's heading is the place it actually lands.
The picture to keep in your head: continuity means you can draw the graph through \( x = a \) without lifting your pen off the paper. Break any one of the three rules and your pen has to jump.
The widget below shows the same neighbourhood three ways. Toggle between continuous, removable hole, and jump, and watch the left limit, the right limit, and \( f(1) \) update — then read the verdict it gives.
Three stories, three outcomes:
- Continuous. Left limit, right limit, and \( f(1) \) are all the same number. Pen stays down. ✓
- Removable hole. Both sides head to the same value, so the limit exists — but \( f(1) \) is missing (or sits at the wrong spot). The destination is fine; the function just isn't standing there. We call it "removable" because filling in that one point would fix everything.
- Jump. The left limit and right limit are different numbers. The two-sided limit doesn't even exist, so there's nothing to match. Your pen has to leap across a gap.
Tip — for "nice" functions, just plug in. Polynomials, and most functions you'll meet built from them, are continuous everywhere. For those, condition 3 is automatic: the limit equals the value. So \( \displaystyle\lim_{x \to 4}(x^2 + 1) \) is simply \( 4^2 + 1 = 17 \) — no table needed. Limits only get interesting when plugging in fails (a \( \tfrac{0}{0} \), a hole, a jump). That's your signal to do real work: factor, simplify, or check the two sides separately.
Worked example 1 — a hidden hole
- Try plugging in \( x = 3 \): top is \( 9 - 9 = 0 \), bottom is \( 3 - 3 = 0 \). That \( \tfrac{0}{0} \) is your cue — the function is undefined here, but a limit may still live here.
- Factor the top as a difference of squares: \( x^2 - 9 = (x - 3)(x + 3) \).
- Cancel the common factor (legal because \( x \neq 3 \) as \( x \to 3 \), so \( x - 3 \neq 0 \)): \( \displaystyle\frac{(x-3)(x+3)}{x-3} = x + 3 \).
- Now the function is just \( x + 3 \), which is continuous, so plug in: \( 3 + 3 = 6 \).
The limit is 6. There's a removable hole in the graph at \( x = 3 \): the function points firmly at 6 but isn't defined there.
Worked example 2 — a genuine jump
- Approach from the left (\( x < 2 \)), so use the rule \( f(x) = x + 1 \): \( \displaystyle\lim_{x \to 2^-} f(x) = 2 + 1 = 3 \).
- Approach from the right (\( x \ge 2 \)), so use the rule \( f(x) = 2x \): \( \displaystyle\lim_{x \to 2^+} f(x) = 2 \cdot 2 = 4 \).
- Compare the two scouts: the left says 3, the right says 4. They disagree.
Because the one-sided limits differ, the two-sided limit does not exist. The graph jumps from height 3 up to height 4 at \( x = 2 \) — your pen has to leap, so \( f \) is not continuous there.
Practice
Try each one yourself, then reveal the full solution.
1. Evaluate \( \displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x - 2} \).
Plugging in \( x = 2 \) gives \( \tfrac{4-4}{2-2} = \tfrac{0}{0} \) — undefined, so factor first. The top is a difference of squares: \( x^2 - 4 = (x-2)(x+2) \).
Cancel the \( x - 2 \) (valid since \( x \neq 2 \) on the approach): the function becomes \( x + 2 \). Now plug in: \( 2 + 2 \).
The limit is 4 (with a removable hole at \( x = 2 \)).
2. Is \( f(x) = x^2 + 1 \) continuous at \( x = 0 \), and what is its value there?
Check the three conditions. (1) \( f(0) = 0^2 + 1 = 1 \) is defined. (2) Since \( f \) is a polynomial it's continuous everywhere, so \( \displaystyle\lim_{x \to 0}(x^2 + 1) = 1 \) exists. (3) The limit equals the value, \( 1 = 1 \).
All three hold, so yes, \( f \) is continuous at \( x = 0 \), and the value there is 1. For a polynomial, finding the limit really is just plugging in.
3. For \( f(x) = \dfrac{x^2 - 1}{x - 1} \), give \( \displaystyle\lim_{x \to 1} f(x) \) and state \( f(1) \).
Factor and cancel: \( \dfrac{(x-1)(x+1)}{x-1} = x + 1 \) for \( x \neq 1 \). So as \( x \to 1 \) the output heads to \( 1 + 1 = 2 \), giving \( \displaystyle\lim_{x \to 1} f(x) = 2 \).
But \( f(1) \) plugs straight into \( \tfrac{0}{0} \), so \( f(1) \) is undefined.
Limit = 2; \( f(1) \) is undefined — a textbook removable hole: the limit exists, but the function isn't defined at the point. This is exactly the function we opened with, and why limits are worth caring about.