Why Boundary Checks Matter
A curve can look plausible near its turning points and still be impossible for the function it is meant to represent. A missing point may have been drawn as an asymptote, a branch may approach the wrong height, or a sketch may include points outside the domain. Boundary checks test these details directly against the function’s limits and any stated derivative information.
For each end of the domain, ask what values of \(x\) can approach it and what the function does there. At a finite boundary, check whether the function is defined at the boundary and evaluate the appropriate one-sided limit. At an unbounded end, check the limit as \(x\) tends to \(+\infty\) or \(-\infty\). These checks concern the behavior required of the graph, not its overall visual appearance.
What to Check at a Boundary
The domain of a function is the set of inputs for which the function is defined. A finite boundary point \(c\) may be included in the domain, excluded from it, or be an excluded point inside the domain’s range of inputs. The graph must show a point at \((c,f(c))\) if \(c\) is in the domain. If \(c\) is excluded, the graph must not include a filled point there just because a limiting value exists.
A one-sided limit describes behavior while approaching from one side. For example, \(\lim_{x\to c^-}f(x)\) uses inputs less than \(c\), and \(\lim_{x\to c^+}f(x)\) uses inputs greater than \(c\). At a boundary of the domain, only the side from which domain points approach is relevant.
These are different situations. A finite limit at an excluded interior point indicates a hole, not a vertical asymptote. If the function is defined at \(c\) but \(f(c)\ne L\), there is a point at \((c,f(c))\) as well as a missing point on the limiting curve at \((c,L)\); this is a removable discontinuity, but not a hole in the strict sense of an undefined function value.
At an unbounded end of the domain, \(\lim_{x\to+\infty}f(x)=L\) or \(\lim_{x\to-\infty}f(x)=L\), for finite \(L\), means that the corresponding end of the graph approaches the horizontal line \(y=L\). The two ends must be checked separately: their limits can differ or one may not be finite.
A Reliable Boundary-Check Procedure
Before accepting a sketch, use a short checklist. The point is not to redraw the whole curve, but to test the features whose location or shape is constrained by the domain and limits.
Mark excluded inputs, finite endpoints, and any directions in which the domain continues without bound.
Use the appropriate one-sided limit on each side that belongs to the domain. Check the actual function value separately when the boundary is included.
A finite limiting value suggests a removable gap; an infinite one-sided limit gives a vertical asymptote on that side.
Compare limits at infinity with any proposed horizontal asymptote. On intervals where derivative information applies, check that the direction of travel agrees.
Worked Example: A Cancelled Factor Produces a Hole
Worked Example: A Cancelled Factor Produces a Hole
Consider \(f(x)=\dfrac{x^2-1}{x-1}\), with its original formula defined only when \(x\ne1\). A proposed sketch shows a vertical asymptote at \(x=1\). Test that feature and the behavior at infinity.
First factor the numerator, while keeping the original domain restriction:
The cancellation is valid only for \(x\ne1\), so it does not put \(1\) back into the domain. To find the limit at the excluded input, use the simplified expression for nearby \(x\):
Both one-sided limits are finite and agree, so \(\lim_{x\to1}f(x)=2\). The graph approaches \((1,2)\), but \(f(1)\) is undefined. Thus \((1,2)\) is a hole, not a vertical asymptote.
The derivative also checks the shape away from the hole. By the quotient rule, for \(x\ne1\),
This agrees with the derivative of \(x+1\), so the graph is increasing on each side of the gap. Finally, because \(f(x)=x+1\) throughout its domain,
Check: The sketch should be the line \(y=x+1\) with only the point \((1,2)\) removed. Substituting \(x=1\) into the original formula makes its denominator zero, confirming that the point is absent. A vertical branch at \(x=1\) would contradict the finite limit.
Worked Example: Two Infinite Limits at a Vertical Asymptote
Worked Example: Two Infinite Limits at a Vertical Asymptote
Consider \(g(x)=\dfrac{1}{x-2}\). A sketch proposes a vertical asymptote at \(x=2\), a horizontal asymptote at \(y=0\), and decreasing branches. Check all three claims.
The denominator is zero at \(x=2\), so the domain is \(\mathbb{R}\setminus\{2\}\). For \(x<2\), the denominator approaches zero through negative values as \(x\to2^-\); its reciprocal becomes arbitrarily negative. For \(x>2\), the denominator approaches zero through positive values. Therefore,
Each side has an infinite one-sided limit, so \(x=2\) is a vertical asymptote. At the unbounded ends, divide by the growing denominator to obtain
Thus \(y=0\) is a horizontal asymptote at both ends. The derivative is
because \((x-2)^2>0\) wherever the derivative is defined. Hence each branch is decreasing on its own interval of definition. The values \(g(1)=1/(1-2)=-1\) and \(g(3)=1/(3-2)=1\) also confirm the signs on the two sides.
Check: The left branch must decrease toward \(-\infty\) as it approaches \(2\), and the right branch must decrease from \(+\infty\) as it moves away from \(2\). Both branches approach \(0\) at their respective infinite ends. These checks agree with the proposed asymptotes and directions.
Worked Example: Included and Excluded Endpoints
Worked Example: Included and Excluded Endpoints
Let \(h(x)=x^2\) on the restricted domain \([-2,1)\). A proposed sketch has a filled point at \((-2,4)\), an open circle at \((1,1)\), and a lowest point at \((0,0)\). Check the boundary markers and the stated shape.
The domain includes \(-2\), so evaluate the function there:
Therefore, \((-2,4)\) belongs to the graph and should be marked with a filled point. The domain excludes \(1\), but values in the domain approach \(1\) from the left. The corresponding limit is
So the graph approaches \((1,1)\) without containing that endpoint; an open circle there is appropriate. This is a finite boundary limit, not a vertical asymptote. There is no end behavior at \(+\infty\) or \(-\infty\) to check, because the domain is bounded.
On the interior of the domain, \(h'(x)=2x\). It is negative for \(-2<x<0\) and positive for \(0<x<1\). Thus the graph decreases toward \(x=0\) from the left and increases to the right. Also, \(h(0)=0\), and \(x^2\ge0\) for every real \(x\), so \((0,0)\) is the lowest point on this domain.
Check: The proposed sketch has the correct endpoint inclusion, endpoint limit, and direction on either side of \(0\). Replacing the open circle at \((1,1)\) with a filled point would incorrectly add \(1\) to the domain.
Why Derivative Information Can Test a Sketch
Limits tell us what a graph approaches; derivative signs can test whether its direction between boundaries is possible. The connection follows from the Mean Value Theorem: if \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then for some \(c\in(a,b)\), \(f(b)-f(a)=f'(c)(b-a)\). For \(a<b\), if \(f'(x)>0\) throughout \((a,b)\), then \(f'(c)>0\), so \(f(b)-f(a)>0\). Therefore \(f(b)>f(a)\). The function is increasing on the interval. If \(f'(x)<0\) throughout, the same equation gives \(f(b)<f(a)\), so the function is decreasing.
The hypotheses matter. Apply this reasoning on intervals where the function is continuous on the closed subinterval and differentiable in its interior. A hole or vertical asymptote can split the domain into separate intervals; monotonicity on one side does not automatically determine behavior on the other.
Common Mistake: Treating Every Excluded Input as an Asymptote
An undefined formula at \(x=c\) does not by itself imply a vertical asymptote. The behavior near \(c\) decides the classification. In the first example, cancelling \(x-1\) left the finite limit \(2\), so the graph had a hole. In the second, the one-sided limits were infinite, so the graph had a vertical asymptote.
Another mistake is to use a two-sided limit at a domain endpoint where the function is available from only one side. For \(h\) on \([-2,1)\), the relevant limit at \(1\) is the left-hand limit. The right-hand side contains no points from the domain and cannot describe the graph’s approach to that endpoint.
Using Checks to Reject an Impossible Sketch
These checks are valuable because a drawing can conceal a contradiction. A curve may seem smooth and well-shaped but still place a filled point outside the domain, approach a finite point where a limit is infinite, or rise on an interval where the derivative is negative. Calculating the boundary limits and comparing them with the sketch turns visual judgment into a precise test.
When a proposed feature fails, identify the exact conflict: the domain, a one-sided limit, an infinite-end limit, or a derivative sign. Then correct only what the evidence requires. A finite limit does not determine the entire curve, for example, but it rules out an infinite branch at that boundary. Likewise, a derivative sign fixes direction on an interval but does not by itself determine the branch’s height or curvature.
Check Your Understanding
For each question, identify the relevant domain or limit check before deciding what a sketch should show.
- If an interior point \(c\) is excluded but both one-sided limits equal the same finite number \(L\), what feature should the sketch show at \((c,L)\)?
- What limit behavior is required for \(x=c\) to be a vertical asymptote on a given side?
- If \(\lim_{x\to+\infty}f(x)=4\), what horizontal line does the graph approach at that end?
- On a restricted domain \([a,b)\), which one-sided limit is relevant at \(b\), and should \((b,f(b))\) be included?
- Suppose \(f'(x)<0\) on an interval where the continuity and differentiability conditions hold. Can a proposed sketch increase from left to right throughout that interval?
