Notes
What a limit means
$\displaystyle \lim_{x \to a} f(x) = L$ means $f(x)$ gets as close as we like to $L$ when $x$ is close enough to $a$ (but not equal to $a$). The value $f(a)$ itself doesn't matter.
- One-sided limits: $\displaystyle \lim_{x \to a^-}$ from the left, $\displaystyle \lim_{x \to a^+}$ from the right. The two-sided limit exists only if both sides agree.
- You can estimate limits from graphs and tables, then confirm them with algebra.
Finding limits with algebra
- Substitute first. If you get a real number, that's the limit (for continuous functions).
- If you get $\displaystyle \frac{0}{0}$, rewrite: factor and cancel, multiply by a conjugate, or simplify a complex fraction.
- Special limits: $\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x} = 0$.
- Squeeze theorem: if $g(x) \le f(x) \le h(x)$ near $a$ and $g$ and $h$ both approach $L$, then so does $f$.
Continuity
$f$ is continuous at $x = a$ when all three are true:
- $f(a)$ is defined
- $\displaystyle \lim_{x \to a} f(x)$ exists
- $\displaystyle \lim_{x \to a} f(x) = f(a)$
Types of discontinuity: removable (a hole), jump (one-sided limits differ), infinite (vertical asymptote).
Limits involving infinity
- $\displaystyle \lim_{x \to a} f(x) = \pm\infty$ means a vertical asymptote at $x = a$.
- $\displaystyle \lim_{x \to \pm\infty} f(x) = L$ means a horizontal asymptote $y = L$. For rational functions, compare degrees.
- Growth order as $x \to \infty$: $\ln x \ll x^n \ll e^x$.
Intermediate Value Theorem (IVT)
If $f$ is continuous on $[a, b]$ and $k$ is between $f(a)$ and $f(b)$, then $f(c) = k$ for at least one $c$ in $[a, b]$.
Topics in this unit
Each topic has its own short notes and practice.
- 1.1Introducing Calculus: Can Change Occur at an Instant?
- 1.2What Is a Limit?
- 1.3Estimating Limit Values from Graphs
- 1.4Estimating Limit Values from Tables
- 1.5Determining Limits Using Algebraic Properties of Limits
- 1.6Determining Limits Using Algebraic Manipulation
- 1.7Selecting Procedures for Determining Limits
- 1.8Determining Limits Using the Squeeze Theorem
- 1.9Connecting Multiple Representations of Limits
- 1.10Exploring Types of Discontinuities
- 1.11Defining Continuity at a Point
- 1.12Confirming Continuity over an Interval
- 1.13Removing Discontinuities
- 1.14Connecting Infinite Limits and Vertical Asymptotes
- 1.15Connecting Limits at Infinity and Horizontal Asymptotes
- 1.16Working with the Intermediate Value Theorem (IVT)
Practice
Factor: $\displaystyle \frac{(x-3)(x+3)}{x-3} = x + 3 \to 6$.
$\displaystyle \frac{\sin 5x}{x} = 5 \cdot \frac{\sin 5x}{5x} \to 5 \cdot 1$.
The limit exists (both sides are 4) but doesn't equal $f(2)$, which is a removable discontinuity.
Equal degrees: the ratio of leading coefficients is $\displaystyle \frac{4}{2}$.
Multiply by the conjugate: $\displaystyle \frac{1}{\sqrt x + 2} \to \frac{1}{4}$.
Match the pieces at $x = 2$: $2k + 1 = 3$, so $k = 1$.
By the IVT, every value between $-2$ and 6, including 0, is reached.
As $x \to 1^+$ the denominator is a small positive number, so the quotient grows without bound.
Free response
Write your answer first, then open the scoring guide.
Free-response question
Let $g(x) = \frac{x^2 + x - 6}{x^2 - 4}$.
(a) Find $\lim_{x \to 2} g(x)$.
Show the answer and scoring guide
$\displaystyle g(x) = \frac{(x + 3)(x - 2)}{(x + 2)(x - 2)} = \frac{x + 3}{x + 2}$ for $x \ne 2$, so the limit is $\displaystyle \frac{5}{4}$.
Check off the points you earned:
(b) Classify the discontinuities of $g$ at $x = 2$ and $x = -2$.
Show the answer and scoring guide
$x = 2$: removable (hole), since the limit exists. $x = -2$: infinite (vertical asymptote), since the numerator $\to 1 \ne 0$ while the denominator $\to 0$.
Check off the points you earned:
(c) Find $\lim_{x \to \infty} g(x)$.
Show the answer and scoring guide
Equal degrees with leading coefficients 1 and 1, so the limit is 1.
Check off the points you earned: