What is a Limit?

An intuitive introduction to the concept of limits in calculus

Understanding Limits Intuitively

A limit describes what happens to a function as the input gets closer and closer to a certain value. Think of it as asking: "Where is this function heading?"

Key Idea: Limits are about the journey, not the destination. We care about where a function is going, not necessarily where it is.

The Limit Notation

We write:

limxaf(x)=L\lim_{x \to a} f(x) = L

This reads: "The limit of f(x) as x approaches a equals L"

What This Means

  • xax \to a: The variable x gets closer and closer to the value a
  • f(x)Lf(x) \to L: The function values get closer and closer to L
  • We don't care about what happens at x = a, only what happens near it

A Simple Example

Consider the function: f(x)=2x+1f(x) = 2x + 1

What is limx3(2x+1)\lim_{x \to 3} (2x + 1)?

As x gets close to 3:

  • When x = 2.9: f(2.9)=2(2.9)+1=6.8f(2.9) = 2(2.9) + 1 = 6.8
  • When x = 2.99: f(2.99)=2(2.99)+1=6.98f(2.99) = 2(2.99) + 1 = 6.98
  • When x = 2.999: f(2.999)=2(2.999)+1=6.998f(2.999) = 2(2.999) + 1 = 6.998

The values are getting closer to 7!

Therefore: limx3(2x+1)=7\lim_{x \to 3} (2x + 1) = 7

Why Limits Matter

Limits are the foundation of calculus because they help us:

  • Define derivatives (rate of change)
  • Define integrals (area under curves)
  • Understand function behavior near problematic points
  • Handle infinity and infinitesimal quantities

Important Note

The limit of f(x) as x approaches a might be different from f(a), or f(a) might not even exist!

📚 Practice Problems

1Problem 1easy

Question:

Using your intuition, estimate limx2(3x1)\lim_{x \to 2} (3x - 1) by checking values near x = 2.

💡 Show Solution

Let's check values approaching 2 from both sides:

From the left (x < 2):

  • x = 1.9: f(1.9)=3(1.9)1=4.7f(1.9) = 3(1.9) - 1 = 4.7
  • x = 1.99: f(1.99)=3(1.99)1=4.97f(1.99) = 3(1.99) - 1 = 4.97
  • x = 1.999: f(1.999)=3(1.999)1=4.997f(1.999) = 3(1.999) - 1 = 4.997

From the right (x > 2):

  • x = 2.1: f(2.1)=3(2.1)1=5.3f(2.1) = 3(2.1) - 1 = 5.3
  • x = 2.01: f(2.01)=3(2.01)1=5.03f(2.01) = 3(2.01) - 1 = 5.03
  • x = 2.001: f(2.001)=3(2.001)1=5.003f(2.001) = 3(2.001) - 1 = 5.003

Both sides approach 5, so: limx2(3x1)=5\lim_{x \to 2} (3x - 1) = 5