CalculusBeginner

What is a Limit?

A limit describes what value a function approaches as its input gets closer and closer to some value — without necessarily reaching it. Limits are the foundation of calculus: every derivative and every integral is defined using a limit.

1 The Intuition Behind Limits

Imagine walking toward a wall. You get closer and closer — halving the distance each step. You approach the wall, but in this thought experiment you never actually touch it. A limit describes that approach: what value are you getting closer to, regardless of whether you arrive?

Consider the function f(x) = (x² − 1)/(x − 1). At x = 1 this is 0/0, which is undefined. But what happens as x gets close to 1? Factoring: (x²−1)/(x−1) = (x+1)(x−1)/(x−1) = x+1 (when x ≠ 1). So as x → 1, f(x) → 2. The function has a "hole" at x=1 but approaches 2 from both sides.

The Key Distinction

The limit of f(x) as x→a describes what f(x) approaches, not what f(a) equals. The function doesn't need to be defined at x=a for the limit to exist. This distinction — approaching vs arriving — is what makes limits powerful.

2 Limit Notation

lim(x→a) f(x) = L reads: "the limit of f(x) as x approaches a equals L."

This means: as x gets arbitrarily close to a (from either side), f(x) gets arbitrarily close to L. The formal ε-δ definition makes "arbitrarily close" precise, but the intuition is sufficient for most applications.

Evaluating a Limit
Find lim(x→3) (x² − 9)/(x − 3)
1Factor the numerator: x² − 9 = (x+3)(x−3)
2Cancel: (x+3)(x−3)/(x−3) = x+3 (when x ≠ 3)
3As x→3: lim(x→3)(x+3) = 3+3 = 6
4The function is undefined at x=3 but approaches 6
Answer: Limit = 6
Direct Substitution
Find lim(x→4) (x² + 2x − 1)
1This function is defined and continuous at x=4
2Direct substitution works: 4² + 2(4) − 1 = 16 + 8 − 1 = 23
Answer: Limit = 23 — when a function is continuous, the limit equals the function value

3 One-Sided Limits

Sometimes a function approaches different values from the left and right. One-sided limits capture this:

lim(x→a⁻) f(x) — limit from the left (x approaches a from values less than a)
lim(x→a⁺) f(x) — limit from the right (x approaches a from values greater than a)

A two-sided limit exists only when both one-sided limits exist and are equal. If lim(x→a⁻) f(x) ≠ lim(x→a⁺) f(x), the limit does not exist (DNE).

Example with absolute value

f(x) = |x|/x. From the right (x→0⁺): positive values divided by positive = +1. From the left (x→0⁻): negative values divided by negative = +1... wait, negative/negative = +1. Actually: from left, x is negative so |x| = −x, giving −x/x = −1. So lim(x→0⁻) = −1 and lim(x→0⁺) = +1. These differ, so lim(x→0) does not exist.

4 Infinite Limits and Asymptotes

When f(x) grows without bound as x→a, we write lim(x→a) f(x) = ∞. This doesn't mean the limit "equals infinity" — infinity isn't a number — it means the function increases without bound.

Example: lim(x→0) 1/x² = ∞. As x approaches 0, 1/x² grows arbitrarily large. The line x=0 is a vertical asymptote.

Limits at infinity: lim(x→∞) f(x) = L means f(x) approaches L as x grows without bound. lim(x→∞) 1/x = 0. The x-axis (y=0) is a horizontal asymptote of y = 1/x.

5 Why Limits Are the Foundation of Calculus

The derivative is defined as a limit: f'(x) = lim(h→0) [f(x+h) − f(x)] / h. This is the slope of the tangent line — the limit of the slope of a secant line as the two points get infinitely close. Without limits, derivatives cannot be defined.

The definite integral is also a limit: the limit of a sum of infinitely many infinitely thin rectangles under a curve. Without limits, integration cannot be defined.

Every theorem in calculus — the chain rule, the mean value theorem, Taylor series — rests on the concept of a limit. Understanding limits deeply makes every subsequent calculus topic easier to grasp.

Practice Problems

Find lim(x→2) (x² − 4)/(x − 2)
Factor: (x+2)(x−2)/(x−2) = x+2. As x→2: 4
Does lim(x→0) 1/x exist?
From the right: 1/x → +∞. From the left: 1/x → −∞. One-sided limits differ, so the limit does not exist.
Find lim(x→∞) (3x² + 1)/(x² − 5)
Divide top and bottom by x²: (3 + 1/x²)/(1 − 5/x²). As x→∞, 1/x²→0. Limit = 3/1 = 3.