Limits constrain nearby graph points
A limit describes the values of a function at inputs close to a point, not necessarily the value of the function at that point. This makes limits useful when reading or checking a graph: the limit tells us which graph behaviors are possible arbitrarily near the input, even if the graph has a hole there or the function is not defined there.
The domain matters. A claim about what happens as \(x\) approaches \(a\) from the left only has content if the domain contains inputs arbitrarily close to \(a\) from the left. If there are no such inputs, a one-sided limit condition can be vacuously satisfied by any proposed limit. We will include the needed domain condition explicitly.
Finite limits and horizontal bands
The condition \(0<|x-a|\) excludes the point \(a\) itself. Consequently, the limit does not specify \(f(a)\), or even require that \(f(a)\) be defined. Its graph interpretation is precise: for every horizontal band \(L-\varepsilon<y<L+\varepsilon\), all graph points with inputs sufficiently close to \(a\), other than \(a\), lie inside that band.
This condition does not say the graph must approach \(L\) from above or from below. It requires only that the values eventually stay within every prescribed band around \(L\).
Worked Example: A finite limit with a missing point
For \(x\ne 2\), let \(f(x)=\dfrac{x^2-4}{x-2}\). Find the limit as \(x\) approaches \(2\), and state what it requires of the graph.
For \(x\ne2\), factor the numerator and cancel the nonzero factor \(x-2\):
Thus, for \(x\in D=\mathbb{R}\setminus\{2\}\),
Given any \(\varepsilon>0\), choose \(\delta=\varepsilon\). If \(x\in D\) and \(0<|x-2|<\delta\), then
Therefore \(\lim_{x\to2}f(x)=4\). The graph agrees with \(y=x+2\) near \(2\), except that \(x=2\) is absent from the domain, so it has a hole at \((2,4)\). The limit is checked by the displayed inequality; it does not assign a value at the hole.
One-sided limits and the domain
A one-sided limit applies the same idea while approaching from only one direction. The direction must be represented in the domain. This requirement prevents an empty set of nearby inputs from making the definition true for every possible \(L\).
If both sides of the domain accumulate at \(a\), then the two-sided finite limit exists with value \(L\) exactly when both one-sided limits exist with value \(L\). This is a local graph criterion, not an assumption that a plotted curve looks smooth.
Given a two-sided \(\delta\), restrict its implication to inputs on either side. For those inputs, \(0<|x-a|<\delta\), so the same \(\varepsilon\)-bound holds.
For a chosen \(\varepsilon>0\), take a left-side \(\delta_-\) and a right-side \(\delta_+\) from the two definitions. Set \(\delta=\min(\delta_-,\delta_+)\). Every domain input with \(0<|x-a|<\delta\) lies on one of those sides and satisfies the required bound.
Both directions use the stated one-sided accumulation conditions. Without them, a side with no domain inputs close to \(a\) would impose no restriction on \(f(x)\), and its limit could not be used to determine a unique value.
Worked Example: Different one-sided limits rule out a two-sided limit
Let \(f(x)=-1\) for \(x<0\) and \(f(x)=1\) for \(x>0\). The function need not be defined at \(0\). Determine the one-sided limits and whether a two-sided limit exists.
For every domain input \(x<0\), \(f(x)=-1\). Thus for every \(\varepsilon>0\), any \(\delta>0\) gives \(|f(x)-(-1)|=0<\varepsilon\) whenever \(x\in D\) and \(0<-x<\delta\). The left-hand limit is \(-1\).
For every domain input \(x>0\), \(f(x)=1\). Thus for every \(\varepsilon>0\), any \(\delta>0\) gives \(|f(x)-1|=0<\varepsilon\) whenever \(x\in D\) and \(0<x<\delta\). The right-hand limit is \(1\). The domain has points arbitrarily close on both sides, but the one-sided values differ. They cannot both equal one common \(L\), so the two-sided limit does not exist.
Graph check: on every sufficiently small left-hand interval, the graph has height \(-1\); on every sufficiently small right-hand interval, it has height \(1\). A sketch showing the same finite limiting height from both sides would contradict these limits.
One-sided infinite limits constrain graph height
An infinite limit is not a finite limit with an unusually large answer. It states that function values eventually exceed every chosen height, or fall below every chosen negative height. As with finite one-sided limits, the domain must contain inputs arbitrarily close from the specified side.
For example, a right-hand limit of \(+\infty\) means that no fixed upper horizontal line can contain all graph points close enough to the right of \(a\): however high the line is, the graph eventually lies above it. It does not require the graph to be monotone, and it says nothing by itself about the other side of \(a\).
Worked Example: Verifying a one-sided limit of positive infinity
For \(x>0\), let \(f(x)=\dfrac{1}{x^2}\). Show that \(\lim_{x\to0^+}f(x)=+\infty\).
Let \(M>0\) be any requested height. Choose \(\delta=\dfrac{1}{\sqrt{M+1}}\), which is positive. If \(0<x<\delta\), then \(x^2<\delta^2=\dfrac{1}{M+1}\). Since both sides are positive, taking reciprocals reverses the inequality:
Thus every \(M>0\) has a corresponding \(\delta>0\), as required. The domain \((0,\infty)\) has points arbitrarily close to \(0\) from the right, so this one-sided limit is meaningful. The graph must eventually lie above each chosen height on the right of \(0\).
Check the direction: as \(x\) approaches \(0\) through positive values, \(x^2\) becomes positive and smaller than \(\delta^2\); its reciprocal therefore becomes larger, not smaller.
Read a limit as a test of a sketch
When a limit is supplied, test a proposed sketch against the definition rather than relying on a general impression of its shape. A finite limit requires all sufficiently nearby graph points to fit inside every prescribed horizontal band. A one-sided infinite limit requires all sufficiently nearby points on that side to lie above, or below, any prescribed height.
Worked Example: Rejecting a sketch that contradicts a finite limit
Suppose a function is defined for inputs on both sides of \(a=1\), and the information states that \(\lim_{x\to1}f(x)=3\). A proposed sketch shows graph points with \(x\) arbitrarily close to \(1\) from the right and \(f(x)>5\). Can that sketch be correct?
Use the finite-limit definition with \(\varepsilon=1\). It requires a \(\delta>0\) such that every domain input satisfying \(0<|x-1|<\delta\) has \(|f(x)-3|<1\). This inequality is equivalent to \(2<f(x)<4\).
The proposed graph has points with \(x\) arbitrarily close to \(1\) from the right and \(f(x)>5\). In particular, within the required \(\delta\)-neighborhood there is such a point. For that point, \(f(x)>5\) gives \(|f(x)-3|>2\), which contradicts \(|f(x)-3|<1\). The sketch cannot represent the stated limit.
Check: any one such point inside the \(\delta\)-neighborhood violates the required band. The contradiction does not depend on the value of \(f(1)\).
What limits do not determine
Limits place strong local restrictions, but they do not determine every feature of a graph. In particular, a finite limit need not equal the function’s value at the point. For example, define \(g(x)=x\) for \(x\ne0\) and \(g(0)=7\). For \(x\ne0\), \(|g(x)-0|=|x|\), so choosing \(\delta=\varepsilon\) proves \(\lim_{x\to0}g(x)=0\), even though \(g(0)=7\).
Similarly, a right-hand limit gives no direct information about values on the left, and an infinite limit from one side does not dictate what happens on the other side. A sketch must obey each stated local constraint, but it may still have many different shapes consistent with that information. The limit alone also does not say whether values approach the limiting height from above or below, or whether the graph is increasing or decreasing.
A practical checklist for graph evidence
Before accepting a local sketch, check the domain and then translate the limit into a constraint on graph points:
- For a finite two-sided limit, check that the domain accumulates on both sides and that nearby values fit every horizontal band around the stated limit.
- For a one-sided finite limit, check only the specified side, but make sure the domain has points arbitrarily close on that side.
- For \(+\infty\), check that values eventually exceed every chosen positive height; for \(-\infty\), check that they eventually lie below every chosen negative height.
- Do not infer the value at the limiting point from the limit. Check whether the function is defined there and, if so, what value it has.
- Do not infer behavior on an unmentioned side from a one-sided limit.
Check Your Understanding
Use the definitions and domain conditions to justify each answer.
- For \(h(x)=\dfrac{x^2-9}{x-3}\), \(x\ne3\), what is \(\lim_{x\to3}h(x)\), and what does it require of the graph near \(x=3\)?
- Why must a one-sided limit definition require domain points arbitrarily close from the specified side?
- If \(\lim_{x\to a^-}f(x)=2\) and \(\lim_{x\to a^+}f(x)=2\), what domain condition is needed to conclude \(\lim_{x\to a}f(x)=2\)?
- What does \(\lim_{x\to a^+}f(x)=-\infty\) require for each chosen \(M>0\)?
- Can a function have \(\lim_{x\to a}f(x)=L\) but \(f(a)\ne L\)? Explain using the definition’s condition on \(x\).
