1.6: The Precise Definition of a Limit
- Page ID
- 121754
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Simplifying Expressions
- Simplifying Radical Expressions
- Solving Inequalities
- Solving Linear Inequalities
- Solving Absolute Value Inequalities
- Absolute Value Functions
- The Concept of Absolute Value
- Logic and Proofs
- Quantifiers and Symbolic Logic
- Direct Proofs
Quantifying Closeness
Recall that the distance between two points \(a\) and \(b\) on a number line is given by |\(a−b\)|. From this point forward, we will constantly be using this interpretation.
- The statement |\(f(x)−L | \lt \epsilon \) may be interpreted as: The distance between \(f(x)\) and \(L\) is less than \(\epsilon\).
- The statement \(0<|x−a| \lt \delta\) may be interpreted as: \(x \neq a\) and the distance between \(x\) and \(a\) is less than \(\delta\).
It is also important to look at the following equivalences for absolute value:
- The statement |\(f(x)−L| \lt \epsilon\) is equivalent to the statement \(L − \epsilon \lt f(x) \lt L + \epsilon\).
- The statement \(0\lt|x−a|\lt\delta\) is equivalent to the statement \(a−\delta\lt x\lt a+\delta\), where \(x\neq a\).
A \(\delta\)-neighborhood centered at \(a\) is defined to be the interval \( (a - \delta, a) \cup (a, a + \delta)\) on the \(x\)-axis.
An \(\epsilon\)-neighborhood centered at \(L\) is defined to be the interval \((L - \epsilon, L + \epsilon)\) on the \(y\)-axis.
The following Interactive Element might help you visualize key characteristics of \(\delta\)- and \(\epsilon\)-neighborhoods.
Interact: Move the sliders to adjust \(x = a\), \(\delta\), \(y = L\), and \(\epsilon\).
Observation: Neighborhoods are symmetric about their centers, and the \(\delta\)-neighborhood never includes its center.
The \(\epsilon\) Gauntlet
Interact: Grab the point on the \(x\)-axis at \(x = 1 + \delta\) and move it right and left.
Observation: We have been given a static \(\epsilon\)-neighborhood about \(y = 6\). Our job is to find a small enough \(\delta\)-neighborhood about \(x = 1\) so that \(f(x)\) is within the \(\epsilon\)-neighborhood for all values of \(x\) in our \(\delta\)-neighborhood.
Consider the limit
\[ \lim_{x \to 3}{\left( 2x + 3 \right)} = 9. \nonumber \]
Find an appropriate \(\delta\) so that all the function values, \(f(x) = 2x + 3\), lie within \(\epsilon = 0.0005\) units of \(y = 9\) whenever \(x\) is within \(\delta\) units of \(x = 3\).
The Precise Definition of a Finite Limit at a Finite Number
Let \(f(x)\) be defined for all \(x \neq a\) over an open interval containing \(a\), and let \(L\) be a real number. We say
\[\lim_{x \to a}f(x)=L \nonumber \]
if for every \( \epsilon >0\), there exists a \( \delta >0\), such that if \(0<|x−a|< \delta \), then \(|f(x)−L|< \epsilon \).
Stated with symbolic logic,
\[\lim_{x \to a}f(x)=L \nonumber \]
means
\[ \forall \epsilon \gt 0, \quad \exists \delta \gt 0 \ni \nonumber \]
\[ 0 \lt |x - a| \lt \delta \implies |f(x) - L| \lt \epsilon. \nonumber \]
| Symbol | Meaning |
|---|---|
| \( \forall \) | "for all" or "for each" or "for every" |
| \( \exists \) | "there exists" or "there is" |
| \( \implies \) | "implies" or "then" |
| \( \ni \) | "such that" or "so that" |
The symbol \( \ni \) is used frequently in symbolic logic for "such that", however, many mathematicians do not use it for that purpose. Check with your instructor to see what logic symbol they prefer (many just avoid the confusion entirely and use "s.t.").
Critically Analyzing the Structure of the Precise Definition of a Finite Limit at a Finite Number
The statement itself involves something called a universal quantifier (for every \( \epsilon >0\), or \( \forall \epsilon \gt 0 \)), an existential quantifier (there exists a \( \delta >0\), or \( \exists \delta \gt 0 \)), and, last, a conditional statement (if \(0<|x−a|< \delta \), then \(|f(x)−L|< \epsilon )\). Table \(\PageIndex{1}\) breaks down the definition and translates each part.
| Definition | Symbolic Logic | Translation |
|---|---|---|
| For every \( \epsilon >0\), | \( \forall \epsilon \gt 0 \) | For every positive distance \( \epsilon \) from \(L\), |
| there exists a \( \delta >0\), | \( \exists \delta \gt 0 \) | there is a positive distance \( \delta \) from \(a\), |
| such that | \( \ni \) | such that |
| if \(0<|x−a|< \delta \), then \(|f(x)−L|< \epsilon \). | \( 0 \lt |x - a| \lt \delta \implies |f(x) - L| \lt \epsilon \) | if \(x\) is within \( \delta \) units of \(a\) and \(x \neq a\), then \(f(x)\) is within \(\epsilon\) units of \(L\). |
Graphically Interpreting the Precise Definition of a Finite Limit at a Finite Number
The Importance of Scratch Work
Scratch work often starts by assuming we have arrived at the consequent. We then work backwards to see how we got to the consequent. We hope this will eventually, with some possible conditions, allow us to work all the way back to the antecedent.
Proving Finite Limits at Finite Numbers (symmetric \( \delta \)-neighborhood)
We will take our time going through this first example, but as we progress, the amount of "discussion" within examples will decrease.
Prove
\[ \displaystyle \lim_{x \to 3}{(2x+3)} = 9 \nonumber. \]
Before moving on to the next example, let's take some lessons from Example \( \PageIndex{3} \).
- We always need to do scratch work when attempting a proof. Not doing so is akin to not looking at the cover of the puzzle box before attempting to piece a puzzle together.
- The scratch work starts with the consequent and works backwards to (hopefully) arrive at the antecedent.
- We have no control over what \( \epsilon \) they hand us, but \( \delta \) will almost always be a function of \( \epsilon \) (the single exception is when proving the limit of a constant function).
- The structure of our proof will be closely aligned with the language of the precise definition of a limit with the exception that we will not be saying, "there exists a \( \delta \gt 0 \)," but rather, "let \( \delta =... \)" because we actually found a workable \( \delta \) in our scratch work.
- We have some flexibility in the language used during the proof. You don't always have to write, "\( \forall \epsilon \gt 0 \)." Instead, you could use language.
Proving Finite Limits at Finite Numbers (non-symmetric \( \delta \)-neighborhoods)
Find \( \delta \gt 0 \) such that if \( 0 \lt \left| x - \frac{\pi}{3} \right| \lt \delta \), then \( \left| 6 \sin{(x)} - 3 \sqrt{3} \right| \lt 0.1 \).
Prove each limit.
- \( \displaystyle \lim_{x \to −4}{(x^2 + 8x + 27)} = 11 \)
- \( \displaystyle \lim_{x \to −1}{(3x^2 + x + 3)} = 5 \)
- \( \displaystyle \lim_{x \to 2}{(x^3 - 1)} = 7 \)
You will find that, in general, the more complex a function, the more likely it is that the algebraic approach is the easiest to apply. The algebraic approach is also more useful in proving statements about limits.
Given
\[ f(x) = \begin{cases}
0, & \text{ if }x \text{ is rational} \\
x^4, & \text{ if }x \text{ is irrational} \end{cases} \nonumber \]
Prove \( \displaystyle \lim_{x \to 0} f(x) = 0 \).
One-Sided Limits
Limit from the Right: Let \(f(x)\) be defined over an open interval of the form \((a,b)\) where \(a<b\). Then
\[\lim_{x \to a^+}f(x)=L \nonumber \]
if for every \( \epsilon >0\), there exists a \( \delta >0\), such that if \(0<x−a< \delta \), then \(|f(x)−L|< \epsilon \).
Limit from the Left: Let \(f(x)\) be defined over an open interval of the form \((b,c)\) where \(b<c\). Then,
\[\lim_{x \to c^−}f(x)=L \nonumber \]
if for every \( \epsilon >0\),there exists a \( \delta >0\) such that if \( − \delta <x−c<0\), then \(|f(x)−L|< \epsilon \).
Prove that
\[\lim_{x \to 2^+}{\left(\sqrt{x−2} - 1\right)} = -1.\nonumber \]


