Introduction to limits, limits on graphs, and limits of infinity.

Before we introduce limits, you can review some notes about discontinuities without limits: lecture 11

What is a limit?

A limit is the value that a function approaches as the input approaches a certain value.

We say that the limit of f(x)f(x) as xx approaches aa is LL if the function f(x)f(x) gets close to LL as xx gets close to aa.

lim⁡x→af(x)=L\lim_{x\to a} f(x) = L

This is read as "the limit of f(x)f(x) as xx approaches aa is LL",
or "the limit of f(x)f(x) as xx goes to aa is LL".

How to find a limit?

  1. Substitute the value of the input into the function.
  2. If the function is defined at the value of the input, then the limit is the value of the function.
  3. If the function is not defined at the value of the input, then the limit is the value that the function approaches as the input approaches the value of the input.

Example for a defined point
The limit of f(x)=x2f(x) = x^2 as xx approaches 2 is 4.

Because f(2)=22=4f(2) = 2^2 = 4.

lim⁡x→2x2=4\lim_{x\to 2} x^2 = 4

Example for an undefined point
The limit of f(x)=(x−1)2x−1f(x) = \frac{(x-1)^2}{x-1} as xx approaches 1 is 0.

Because as xx gets closer to 1, f(x)f(x) gets closer to 0.

lim⁡x→1(x−1)2x−1=lim⁡x→1(x−1)=0\lim_{x\to 1} \frac{(x-1)^2}{x-1} = \lim_{x\to 1} (x-1) = 0

Limits at a point

Limits at a point are the values that a function approaches as the input approaches a certain value.

  1. If the function is defined at the point, then the limit is the value of the function at the point.

  2. If the function is not defined at the point, then the limit is the value that the function approaches as the input approaches the point. This includes holes, jumps, and vertical asymptotes.

Example
If a hole exists at x=3x=3, and f(x)=x2−9x−3f(x)=\frac{x^2-9}{x-3}, then the limit of f(x)f(x) as xx approaches 3 is 6.

Because f(x)=(x−3)(x+3)x−3=x+3f(x)=\frac{(x-3)(x+3)}{x-3}=x+3 for x≠3x\neq 3.

lim⁡x→3x2−9x−3=lim⁡x→3(x+3)=6\lim_{x\to 3} \frac{x^2-9}{x-3} = \lim_{x\to 3} (x+3) = 6

Limits at infinity

Limits at infinity are the values that a function approaches as the input approaches infinity.

Example
The limit of f(x)=x2f(x) = x^2 as xx approaches infinity is infinity.

lim⁡x→∞x2=∞\lim_{x\to \infty} x^2 = \infty

and the limit of f(x)=x2f(x) = x^2 as xx approaches negative infinity is infinity.

lim⁡x→−∞x2=∞\lim_{x\to -\infty} x^2 = \infty

Numerical examples

Let's explore limits numerically by examining how a function behaves as we get closer to a specific point.

Example 1: Simple substitution

Consider the function f(x)=2x+1f(x) = 2x + 1 and find lim⁡x→3f(x)\lim_{x \to 3} f(x).

xx f(x)=2x+1f(x) = 2x + 1
2.9 6.8
2.99 6.98
2.999 6.998
2.9999 6.9998
3 7
3.0001 7.0002
3.001 7.002
3.01 7.02
3.1 7.2

As xx gets closer to 3, f(x)f(x) gets closer to 7. Therefore, lim⁡x→3(2x+1)=7\lim_{x \to 3} (2x + 1) = 7.

Example 2: Function with a hole

Consider the function f(x)=x2−4x−2f(x) = \frac{x^2 - 4}{x - 2} and find lim⁡x→2f(x)\lim_{x \to 2} f(x).

xx f(x)=x2−4x−2f(x) = \frac{x^2 - 4}{x - 2}
1.9 3.9
1.99 3.99
1.999 3.999
1.9999 3.9999
2 undefined
2.0001 4.0001
2.001 4.001
2.01 4.01
2.1 4.1

Even though f(2)f(2) is undefined (division by zero), as xx approaches 2 from both sides, f(x)f(x) approaches 4. Therefore, lim⁡x→2x2−4x−2=4\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4.

Note: We can verify this algebraically:

x2−4x−2=(x−2)(x+2)x−2=x+2 for x≠2\frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2} = x+2 \text{ for } x \neq 2

So lim⁡x→2x2−4x−2=lim⁡x→2(x+2)=4\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} (x+2) = 4

Example 3: One-sided limits

Consider the function f(x)=∣x∣xf(x) = \frac{|x|}{x} and find lim⁡x→0f(x)\lim_{x \to 0} f(x).

xx (approaching from left) f(x)=∣x∣xf(x) = \frac{|x|}{x}
-0.1 -1
-0.01 -1
-0.001 -1
-0.0001 -1
xx (approaching from right) f(x)=∣x∣xf(x) = \frac{|x|}{x}
0.0001 1
0.001 1
0.01 1
0.1 1

From the left: lim⁡x→0−∣x∣x=−1\lim_{x \to 0^-} \frac{|x|}{x} = -1
From the right: lim⁡x→0+∣x∣x=1\lim_{x \to 0^+} \frac{|x|}{x} = 1

Since the left and right limits are different, lim⁡x→0∣x∣x\lim_{x \to 0} \frac{|x|}{x} does not exist.

Exercises

1. Check your understanding

a) What is lim⁡x→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}?





b) What is lim⁡x→5(3x−2)\lim_{x \to 5} (3x-2)?





c) What is lim⁡x→−1∣x∣x\lim_{x \to -1} \frac{|x|}{x}?





d) What is lim⁡x→−1x2−1x−1\lim_{x \to -1} \frac{x^2-1}{x-1}?





2. Limits at infinity

Write 98765 if the limit is ∞\infty and -98765 if the limit is −∞-\infty.

a) What is lim⁡x→∞1x\lim_{x \to \infty} \frac{1}{x}?





b) What is lim⁡x→−∞1x\lim_{x \to -\infty} \frac{1}{x}?





c) What is lim⁡x→∞x2x+1\lim_{x \to \infty} \frac{x^2}{x+1}?





d) What is lim⁡x→−∞x2x+1\lim_{x \to -\infty} \frac{x^2}{x+1}?





e) What is lim⁡x→∞2x\lim_{x \to \infty} 2^x?





f) What is lim⁡x→−∞2x\lim_{x \to -\infty} 2^x?





g) What is lim⁡x→∞log⁡2(x)\lim_{x \to \infty} \log_2(x)?





h) What is lim⁡x→0+log⁡2(x)\lim_{x \to 0^+} \log_2(x)?





i) What is lim⁡x→∞log⁡2(1x)\lim_{x \to \infty} \log_2(\frac{1}{x})?





j) What is lim⁡x→∞2x+4x−2\lim_{x \to \infty} \frac{2x+4}{x-2}?





k) What is lim⁡x→−∞2x+4x−2\lim_{x \to -\infty} \frac{2x+4}{x-2}?





3. Graph

Using graphical software (like Desmos or GeoGebra), plot the following functions and identify the limits:

a) f(x)=x3−x2−x+1x−1f(x) = \frac{x^3-x^2-x+1}{x-1} as xx approaches 1.

b) g(x)=3x−1xg(x) = \frac{3^x-1}{x} as xx approaches 0.

c) h(x)=1+x−1xh(x) = \frac{\sqrt{1+x}-1}{x} as xx approaches 0.

d) k(x)=log⁡2(x+1)xk(x) = \frac{\log_2(x+1)}{x} as xx approaches 0. (Compare log⁡2\log_2 when it is replaced with ln⁡\ln)

4. Compound limits

Evaluate the following limits:

a) lim⁡x→∞log⁡3(x3−8x+1)\lim_{x \to \infty} \log_3(\frac{x^3-8}{x+1})

b) lim⁡x→∞(x2+5−x)\lim_{x \to \infty} (\sqrt{x^2+5} - x)

c) lim⁡x→−∞x2+1x\lim_{x \to -\infty} \frac{\sqrt{x^2+1}}{x}

d) lim⁡x→2x+7−3x−2\lim_{x \to 2} \frac{\sqrt{x+7} - 3}{x-2}