M171 Calculus I · Unit 1: Understanding the derivative · on Exam 1
§1.2 The notion of limit
Study notes for The notion of limit in M171 Calculus I at Montana State University: the big ideas, key formulas, a worked example, common mistakes and an exam tip. The same topic has endless practice questions and flashcards in Mathub.
Big ideas
- $\displaystyle\lim_{x\to a} f(x) = L$ means $f(x)$ can be made as close to $L$ as we like by taking $x$ close enough to $a$ (but $x\neq a$). The value $f(a)$ is irrelevant: it may not exist, or may differ from $L$.
- The limit exists exactly when the left-hand limit $\lim_{x\to a^-}f(x)$ and right-hand limit $\lim_{x\to a^+}f(x)$ both exist and are equal.
- Three kinds of "plug-in" outcomes. If substituting gives $\frac{c}{d}$ with $d\neq0$: that is the limit (a valid answer). If it gives $\frac{c}{0}$ with $c\neq0$: vertical asymptote, the limit does not exist (it blows up to $\pm\infty$). If it gives $\frac{0}{0}$: indeterminate form, do more algebra.
- Algebra tools for $\frac00$: factor and cancel; multiply by a conjugate; combine fractions with a common denominator.
- Estimate limits from a table (approach from both sides) or a graph (follow the curve toward $x=a$ from each side).
Key formulas
Two-sided limit
$$\lim_{x\to a} f(x) = L \iff \lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = L$$
Factor and cancel
$$\lim_{x\to 2}\frac{x^2+3x-10}{x-2} = \lim_{x\to 2}\frac{(x+5)(x-2)}{x-2} = 7$$
Conjugate
$$\lim_{x\to 2}\frac{\sqrt{x+7}-3}{x-2} = \lim_{x\to 2}\frac{1}{\sqrt{x+7}+3} = \frac16$$
Common denominator
$$\lim_{x\to 4}\frac{\frac1x-\frac14}{x-4} = \lim_{x\to 4}\frac{-1}{4x} = -\frac1{16}$$
Worked example
Classify the form and evaluate: (a) $\lim_{x\to0}\frac{x+4}{x}$, (b) $\lim_{x\to0}\frac{\sin x}{\cos x}$, (c) $\lim_{x\to1}\frac{x^2-1}{x-1}$.
Show the solution
(a) Substituting gives $\frac40$: vertical asymptote, the limit does not exist. (b) Gives $\frac01 = 0$: a valid answer, the limit is $0$. (c) Gives $\frac00$: indeterminate. Factor: $\frac{(x-1)(x+1)}{x-1} = x+1 \to 2$.
Common mistakes
- Saying "$\frac00 = 0$" or "$\frac00$ is undefined so the limit does not exist." $\frac00$ means keep working.
- Confusing $\lim_{x\to a}f(x)$ with $f(a)$. A hole in the graph still has a limit.
- Only checking one side of a piecewise function.
Exam tipOn the exam, name the form first ("this is $\frac00$, indeterminate"), then show the algebra. Naming the form earns rubric points even if the algebra slips.
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