Why Endpoints Need a Different Test
A function whose domain begins at \(x=0\) has no domain values to the left of \(0\). To test continuity there, it would make no sense to require the function to approach its value from both sides using points in the domain. The relevant question is whether values of the function approach \(f(0)\) as inputs approach \(0\) from within the domain.
This is why continuity at an endpoint uses a one-sided limit. At an interior point, by contrast, inputs in the domain are available on both sides, so continuity requires the two-sided limit to equal the function value.
One-Sided Continuity at an Endpoint
Suppose a function \(f\) is defined on an interval that includes its left endpoint \(a\), such as \([a,b]\) or \([a,\infty)\). Continuity at \(a\) is determined by the right-hand limit. If \(b\) is an included right endpoint, continuity at \(b\) is determined by the left-hand limit.
The superscript \(+\) means that \(x\) approaches \(a\) with \(x>a\); the superscript \(-\) means that \(x\) approaches \(b\) with \(x<b\). The endpoint must itself belong to the domain, because continuity at a point requires the function to have a value there. If an interval is open at an endpoint, that endpoint is not a point of the function's domain, so continuity of the function at that point is not defined.
The condition has the same three ingredients that occur in the definition of continuity at a point: the function value exists, the appropriate limit exists, and the limit equals the value. At an endpoint, the appropriate limit is one-sided rather than two-sided.
Why the One-Sided Limit Is the Right Test
For a left endpoint \(a\), the definition of the right-hand limit says that \(f(x)\) approaches a number \(L\) as \(x\) approaches \(a\) from the right if, for every \(\varepsilon>0\), there is a \(\delta>0\) such that \(x\) in the domain and \(0<x-a<\delta\) imply \(|f(x)-L|<\varepsilon\). Continuity at \(a\) asks for that limiting number to be \(f(a)\).
Here \(\operatorname{dom}(f)\) denotes the domain of \(f\). Including equality in the distance condition allows \(x=a\) (or \(x=b\)); at that input, the difference from the endpoint value is exactly zero. The one-sided limit and the endpoint condition are equivalent: the limit guarantees closeness to \(f(a)\) (or \(f(b)\)) for all sufficiently nearby domain inputs on the permitted side, and the displayed condition states precisely that guarantee.
At an interior point \(c\), there are domain inputs both below and above \(c\). Continuity there requires $$ \lim_{x\to c} f(x)=f(c), $$ so the values must approach \(f(c)\) from both sides. A right-hand limit alone is not enough at an interior point.
Worked Endpoint Tests
Worked Example: Continuity of a Square Root at Its Left Endpoint
Let \(f(x)=\sqrt{x}\) with domain \([0,\infty)\). Determine whether \(f\) is continuous at \(0\).
The point \(0\) is in the domain and is its left endpoint. We therefore use the right-hand limit. Since \(\sqrt{x}\) approaches \(0\) as \(x\) approaches \(0\) from above,
The endpoint value is
For a direct check, let \(\varepsilon>0\) and choose \(\delta=\varepsilon^2\). If \(x\in[0,\infty)\) and \(0\le x<\delta\), then \(0\le\sqrt{x}<\varepsilon\), so \(|f(x)-f(0)|=|\sqrt{x}-0|<\varepsilon\). Thus the right-hand limit equals the endpoint value, and \(f\) is continuous at \(0\).
Check: Substituting the endpoint into the function gives \(f(0)=0\), exactly the value of the right-hand limit.
Worked Example: Continuity at a Right Endpoint
Let \(g(x)=x^2+1\) on the domain \([1,3]\). Test continuity at the right endpoint \(3\).
Since \(3\) is the right endpoint, inputs approach it from the left. The polynomial expression has limit
The value at the endpoint is
Therefore \(\lim_{x\to 3^-}g(x)=g(3)\), and \(g\) is continuous at \(3\) relative to its domain \([1,3]\). No right-hand limit is needed for this endpoint test, because inputs greater than \(3\) are not in the domain.
Check: The left-hand limit gives \(10\), and substituting \(3\) into the formula also gives \(10\).
Worked Example: An Endpoint Where Continuity Fails
Define \(h\) on \([0,\infty)\) by \(h(0)=0\) and \(h(x)=1\) for \(x>0\). Decide whether \(h\) is continuous at \(0\).
The domain begins at \(0\), so calculate the right-hand limit. For every \(x>0\), \(h(x)=1\), hence
But the value assigned at the endpoint is \(h(0)=0\). Therefore
so \(h\) is not continuous at \(0\). The mismatch is between the values approached from inside the domain and the value at the endpoint.
Check: Every positive input, however close to \(0\), gives output \(1\), while the endpoint itself gives output \(0\); the limit and endpoint value are indeed unequal.
Endpoint Versus Interior: A Worked Comparison
Worked Example: A One-Sided Match Is Not Enough at an Interior Point
Define \(q\) on \([-1,1]\) by \(q(x)=1\) when \(x<0\), and \(q(x)=2\) when \(x\ge 0\). Compare the test at \(0\) with an endpoint test.
First, \(0\) is an interior point of \([-1,1]\). The left-hand limit is \(1\), because \(q(x)=1\) for \(x<0\). The right-hand limit is \(2\), because \(q(x)=2\) for \(x>0\). Thus
These one-sided limits are unequal, so the two-sided limit at \(0\) does not exist. Although \(q(0)=2\), \(q\) is not continuous at this interior point.
Now test the left endpoint \(-1\). The appropriate limit is from the right. For \(x\) just greater than \(-1\), \(q(x)=1\), so
Thus \(q\) is continuous at \(-1\) relative to \([-1,1]\). The calculation at the endpoint uses the side inside the domain; the calculation at \(0\) needs both sides.
Check: At \(-1\), the right-hand limit equals the endpoint value \(1\). At \(0\), the left and right limits are \(1\) and \(2\), so they cannot form a two-sided limit equal to \(q(0)=2\).
A Common Mistake: Requiring Two Sides at an Endpoint
A frequent error is to reject continuity at a domain endpoint because values on the other side do not approach the endpoint value—or because the function is not even defined there. For example, for \(f(x)=\sqrt{x}\) on \([0,\infty)\), one might write, “There is no real value of \(\sqrt{x}\) for \(x<0\), so the two-sided limit at \(0\) does not exist; therefore \(f\) is discontinuous at \(0\).” The conclusion is wrong: it applies a two-sided test that the endpoint does not require.
The reverse mistake is just as important: using only a right-hand or left-hand limit at an interior point. The example \(q\) shows why that fails. At \(0\), the right-hand limit equals \(q(0)\), but the left-hand limit differs, so the two-sided limit does not exist.
A Reliable Endpoint Procedure
Continuity at a point is a question about a point in the domain. If the endpoint is excluded, the function has no value there to compare with a limit.
At a left endpoint, approach from the right; at a right endpoint, approach from the left.
Use the formula or definition that applies to domain inputs near the endpoint, without evaluating outside the domain.
The function is continuous there exactly when the one-sided limit exists and equals the function value.
This endpoint test lets us discuss continuity on domains such as \([a,b]\), \([a,\infty)\), and \((-\infty,b]\) without inventing function values outside those domains. The direction of approach is determined by the domain, not by a preference for left-hand or right-hand limits.
Check Your Understanding
For each question, identify the relevant limit before deciding whether continuity holds.
- For \(f(x)=x^3\) on \([0,4]\), which one-sided limit tests continuity at \(0\)?
- For \(g(x)=\sqrt{5-x}\) on \([1,5]\), write the continuity condition at the right endpoint.
- A function has domain \((2,6]\). Is continuity of that function at \(2\) defined? Explain.
- At an interior point \(c\), the right-hand limit equals \(f(c)\), but the left-hand limit is different. Is \(f\) continuous at \(c\)?
- For \(h(0)=3\) and \(h(x)=x+1\) for \(x>0\), with domain \([0,\infty)\), is \(h\) continuous at \(0\)? Compare the right-hand limit with \(h(0)\).
