Notera

Lesson 1: Limits & Continuity

What You'll Learn

In this lesson you will study the concept of a limit — the foundation of all of calculus. You'll learn how to evaluate limits graphically, numerically, and algebraically, explore one-sided limits, and understand what it means for a function to be continuous.
Definition

Limit of a Function

We write

lim⁡x→af(x)=L\lim_{x \to a} f(x) = L

and say "the limit of f(x)f(x) as xx approaches aa is LL" if the values of f(x)f(x) can be made arbitrarily close to LL by taking xx sufficiently close to aa (from either side), without letting x=ax = a.

The limit describes the tendency of a function near a point — even if the function is undefined there.
On the graph of y=f(x)y = f(x), the limit as x→ax \to a equals LL when the curve approaches the height LL from both the left and the right as xx gets close to aa. A hole at (a,f(a))(a, f(a)) does not affect the limit.
Definition

One-Sided Limits

The left-hand limit is:
lim⁡x→a−f(x)=L\lim_{x \to a^-} f(x) = L
(approaching aa from values less than aa).

The right-hand limit is:
lim⁡x→a+f(x)=L\lim_{x \to a^+} f(x) = L
(approaching aa from values greater than aa).

The two-sided limit lim⁡x→af(x)\lim_{x \to a} f(x) exists if and only if both one-sided limits exist and are equal.
Suppose lim⁡x→af(x)=L\lim_{x\to a}f(x) = L and lim⁡x→ag(x)=M\lim_{x\to a}g(x) = M. Then:

1. Sum: lim⁡x→a[f(x)+g(x)]=L+M\lim_{x\to a}[f(x)+g(x)] = L + M
2. Difference: lim⁡x→a[f(x)−g(x)]=L−M\lim_{x\to a}[f(x)-g(x)] = L - M
3. Constant multiple: lim⁡x→a[cf(x)]=cL\lim_{x\to a}[cf(x)] = cL
4. Product: lim⁡x→a[f(x)⋅g(x)]=L⋅M\lim_{x\to a}[f(x)\cdot g(x)] = L \cdot M
5. Quotient: lim⁡x→af(x)g(x)=LM\lim_{x\to a}\frac{f(x)}{g(x)} = \frac{L}{M}, provided M≠0M \neq 0
6. Power: lim⁡x→a[f(x)]n=Ln\lim_{x\to a}[f(x)]^n = L^n for any positive integer nn
You can break a complicated limit into simpler pieces using these algebraic rules.
Ex

Example — Evaluating a Limit by Factoring

Find lim⁡x→3x2−9x−3\lim_{x \to 3} \frac{x^2 - 9}{x - 3}.
Solution Steps
If g(x)≤f(x)≤h(x)g(x) \leq f(x) \leq h(x) for all xx near aa (except possibly at aa), and

lim⁡x→ag(x)=lim⁡x→ah(x)=L\lim_{x\to a}g(x) = \lim_{x\to a}h(x) = L

then lim⁡x→af(x)=L\lim_{x\to a}f(x) = L as well.

Classic application: lim⁡x→0sin⁡xx=1\lim_{x\to 0}\frac{\sin x}{x} = 1, proved by squeezing between cos⁡x\cos x and 11.
If a function is trapped between two others that share the same limit, it must have that limit too.
Ex

Example — A Famous Limit

Show that lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1.
Solution Steps
Definition

Continuity

A function ff is continuous at x=ax = a if all three conditions hold:

1. f(a)f(a) is defined.
2. lim⁡x→af(x)\lim_{x \to a} f(x) exists.
3. lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a).

Intuitively, the graph has no break, jump, or hole at x=ax = a.

Polynomials, rational functions (on their domain), sin⁡x\sin x, cos⁡x\cos x, exe^x, and ln⁡x\ln x are all continuous on their domains.
If ff is continuous on [a,b][a, b] and NN is any number between f(a)f(a) and f(b)f(b), then there exists at least one c∈(a,b)c \in (a,b) such that f(c)=Nf(c) = N.

In plain language: a continuous function that goes from one value to another must hit every value in between. This is often used to show that an equation has a solution in an interval.
Continuous functions can't 'skip' values. If f(a) < 0 and f(b) > 0, there must be a root between a and b.
Ex

Example — Using the IVT

Show that x3−x−1=0x^3 - x - 1 = 0 has a solution between x=1x = 1 and x=2x = 2.
Solution Steps
Definition

Infinite Limits & Limits at Infinity

Infinite limit: lim⁡x→af(x)=∞\lim_{x \to a} f(x) = \infty means f(x)f(x) grows without bound as x→ax \to a. The line x=ax = a is a vertical asymptote.

Limit at infinity: lim⁡x→∞f(x)=L\lim_{x \to \infty} f(x) = L means f(x)→Lf(x) \to L as xx grows without bound. The line y=Ly = L is a horizontal asymptote.

For rational functions p(x)q(x)\frac{p(x)}{q(x)}:
- If deg⁡p<deg⁡q\deg p < \deg q: horizontal asymptote y=0y = 0.
- If deg⁡p=deg⁡q\deg p = \deg q: horizontal asymptote y=leading coeff of pleading coeff of qy = \frac{\text{leading coeff of }p}{\text{leading coeff of }q}.
- If deg⁡p>deg⁡q\deg p > \deg q: no horizontal asymptote.
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Practice Problem
easy
Evaluate lim⁡x→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}.
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Practice Problem
medium
Evaluate lim⁡x→0sin⁡5xx\lim_{x \to 0} \frac{\sin 5x}{x}.
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Practice Problem
medium
At which value(s) of xx is f(x)=x+1x2−1f(x) = \frac{x+1}{x^2-1} discontinuous?
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Practice Problem
easy
Evaluate lim⁡x→∞3x2+25x2−1\lim_{x \to \infty} \frac{3x^2 + 2}{5x^2 - 1}.
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