How Math Works
Analysis & Calculusconceptintermediate

Limits

A limit measures the approach, not the arrival.

The formula

lim(x→a) f(x) = L

How to read it: the value L that f(x) heads toward as x is pushed arbitrarily close to a

x→a
x approaches a (getting arbitrarily close, never equal)
f(x)
the value of the function there
L
the value f(x) closes in on (the limit)

The hook

A limit doesn't ask 'what is the value there' — it asks 'what does it head toward on the way there'.

In plain words

Push x closer and closer to some value a, and watch which number the function crowds toward.

The intuition

You can name the destination just by watching the approach, even if you never actually arrive. Walking through fog, if every signpost points to one spot, that spot is the limit.

How it's built

What matters is the behavior near x=a, not the value at x=a. Even if that point is a hole, if both sides close in on the same number, the limit exists.

Example

(x²−1)/(x−1) is 0/0 at x=1, so it's undefined there. But it simplifies to x+1, and pushing x toward 1 drives the value to 2 → the limit is 2.

Common mistake

The limit and the actual value can differ. At x=1 the function isn't even defined, yet the limit is a perfectly clean 2.

Where it's used

Derivatives, integrals, continuity, infinite series — all of calculus stands on this single idea of a limit.

Where it came from

In the 1600s people hand-waved about 'infinitely small quantities'. In the 1800s Cauchy and Weierstrass pinned the limit down rigorously, giving calculus its foundation.

Prerequisites

Quick check

What is lim(x→3) (x²−9)/(x−3)?

  • 0
  • 3
  • 6
  • 9

Practice

Find lim(x→2) (x²−4)/(x−2).

Answer: 4

Solution:
  1. Factor the top: x²−4 = (x−2)(x+2)
  2. Cancel → x+2 (for x≠2)
  3. Plug in x=2 → 4

Takeaway: When you hit 0/0, cancel to patch the hole, then substitute.

Find lim(x→0) (3x+1).

Answer: 1

Solution:
  1. This function is smooth everywhere
  2. Just substitute x=0 → 1

Takeaway: No hole means the limit is just the substituted value.

Does lim(x→0) 1/x exist? On your notepad, write what happens from the left and from the right.

Answer: undefined

Solution:
  1. From the right (x→0⁺): 1/x → +∞
  2. From the left (x→0⁻): 1/x → −∞
  3. The two sides disagree, so the limit does not exist

Takeaway: If the two sides head to different values, there is no limit.

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