Three Questions Before You Declare Continuity
Continuity tests can look like short calculations, but a correct conclusion depends on checking the right things in the right order. A limit may exist even when the function is not defined at the point. A function may have a value there that does not match the limit. And at an interior point, checking just one side can miss a mismatch on the other side.
Keep three questions separate: Is the function defined at the point? Do nearby function values approach a limit? If so, does that limit equal the function’s value?
This definition gives a useful checklist, but the domain determines which test is appropriate. Continuity at a point can only be discussed if that point belongs to the function’s domain. At an endpoint of an interval, continuity is tested from within the domain, using the available one-sided limit. At an interior point, values from both sides must agree.
Worked Example: The Limit Exists, but the Assigned Value Is Different
Worked Example: The Limit Exists, but the Assigned Value Is Different
Define \(f\) by \(f(x)=\frac{x^2-1}{x-1}\) for \(x\ne 1\), and \(f(1)=5\). Is \(f\) continuous at \(1\)?
First, the function is defined at \(1\), with \(f(1)=5\). To find the limit, factor the numerator for \(x\ne 1\):
Therefore, as \(x\) approaches \(1\) through inputs other than \(1\), \(f(x)=x+1\), so
Compare the limit with the assigned value: \(2\ne 5\). Hence \(f\) is not continuous at \(1\). The limit calculation describes values near \(1\); it does not change the separately assigned value \(f(1)=5\).
Check: The continuity condition would require \(f(1)=\lim_{x\to1}f(x)\). Substitution of the computed quantities gives \(5=2\), which is false. Thus the conclusion is consistent with the definition.
Error 1: Treating the Function Value as Its Limit
The value \(f(c)\) and the limit \(\lim_{x\to c}f(x)\) answer different questions. The function value asks what the function assigns at exactly \(c\). The limit asks what values \(f(x)\) approach as \(x\) gets close to \(c\), with \(x\ne c\) during the approach.
When \(f\) is continuous at \(c\), these quantities are equal. But their equality is a condition to check, not a fact to assume. In the example above, the function was defined at \(1\) and its limit existed, yet the two numbers disagreed.
Error 2: Ignoring an Input Outside the Domain
A limit may exist at a number that is not in the function’s domain. That fact does not make the function continuous at that number: continuity requires the function to have a value there in the first place. Be especially alert to denominators that become zero, square roots whose radicands are negative, and other restrictions in a formula’s domain.
Worked Example: A Hole in the Domain
Let \(g(x)=\frac{x^2-4}{x-2}\), with domain \(\mathbb{R}\setminus\{2\}\). What can be said about continuity at \(2\)?
Since \(2\) is excluded from the domain, \(g(2)\) is not defined. For \(x\ne 2\), factor and cancel:
The cancellation is valid for \(x\ne2\), which includes every input in the domain. Consequently,
The limit exists, but \(g(2)\) does not. Therefore the original function is not continuous at \(2\), because \(2\) is not in its domain. It is accurate to say that the limit exists there; it is not accurate to say that the original function is continuous there.
Check: The formula \(x+2\), if used to define a new function at every real number, would assign the value \(4\) at \(2\). That is a possible continuous extension, but it is not the original function \(g\), whose domain excludes \(2\).
For continuity on a domain \(D\), the point being tested must belong to \(D\). A continuity claim at a missing input skips a necessary condition, even if the limit calculation is correct.
Error 3: Checking Only One Side at an Interior Point
At an interior point of an interval, inputs can approach from the left and from the right. A two-sided limit exists only if both one-sided limits exist as finite numbers and are equal. To establish continuity, their common value must also equal the function value.
Here is why the two-sided test requires both sides. In the \(\varepsilon\)-\(\delta\) definition, a two-sided limit of \(L\) means that, for every \(\varepsilon>0\), there is a \(\delta>0\) such that \(0<|x-c|<\delta\) implies \(|f(x)-L|<\varepsilon\). This condition applies to inputs on either side of \(c\). Restricting those same inputs to \(c-\delta<x<c\) gives the left-hand limit condition; restricting them to \(c<x<c+\delta\) gives the right-hand condition.
Conversely, suppose both one-sided limits equal \(L\). Given any \(\varepsilon>0\), the left-hand limit provides a positive distance \(\delta_-\) that makes \(|f(x)-L|<\varepsilon\) to the left of \(c\). The right-hand limit provides a positive distance \(\delta_+\) that works to the right. Set \(\delta=\min(\delta_-,\delta_+)\). Then every \(x\) with \(0<|x-c|<\delta\) lies on one of those two sides, and on that side \(|f(x)-L|<\varepsilon\). This is precisely the two-sided limit condition. The two one-sided limits must therefore agree to give a two-sided limit.
Worked Example: The Right Side Matches, but the Left Side Does Not
Define \(h(x)=x\) for \(x<0\) and \(h(x)=x+1\) for \(x\ge0\). Is \(h\) continuous at the interior point \(0\)?
At \(0\), the second formula applies, so \(h(0)=0+1=1\). From the left, \(h(x)=x\), giving
From the right, \(h(x)=x+1\), giving
The one-sided limits disagree: \(0\ne1\). Hence the two-sided limit does not exist, and \(h\) is not continuous at \(0\). The fact that the right-hand limit equals \(h(0)=1\) does not settle the test.
Check: If the right-hand side alone were used, it would give \(1=1\) and suggest continuity. Including the left side gives \(0\ne1\), so that conclusion is incorrect.
Endpoints Require the Available Side
At an endpoint of the function’s domain, there may be inputs on only one side. In that setting, continuity is tested using the one-sided limit from within the domain. Requiring a limit from the other side would ask about inputs for which the function is not defined and would use the wrong test.
Worked Example: Continuity at a Domain Endpoint
Let \(p(x)=\sqrt{x}\) with domain \([0,\infty)\). Is \(p\) continuous at the endpoint \(0\)?
The function is defined there, and \(p(0)=\sqrt{0}=0\). Since the domain has no negative inputs, test the limit from the right:
Therefore \(p\) is continuous at \(0\) relative to its domain. A two-sided test is not required at this endpoint.
Check: The right-hand limit and the function value are both \(0\), so the endpoint continuity condition is satisfied.
A Reliable Continuity Check
Use this sequence to avoid mixing up the three parts of the test:
Confirm that the point \(c\) belongs to the domain, so \(f(c)\) is defined.
At an interior point, examine both one-sided limits. At an endpoint, use the side within the domain.
Verify that the required limit exists as a finite number and equals \(f(c)\).
Each step rules out a different error. A missing function value fails the domain check. Unequal one-sided limits fail the two-sided limit test. A limit different from the function value fails the final comparison.
Why These Distinctions Matter
A continuity conclusion is only as reliable as the conditions used to reach it. If an argument checks only the function value, it says nothing about nearby behavior. If it computes a limit but ignores the domain, it may attribute continuity to a function at an input where the function has no value. If it checks only one side at an interior point, it may miss a jump.
Before using a continuity result, make the tested point and domain explicit, identify whether it is an interior point or an endpoint, and record the value and relevant limit separately. These habits are useful beyond individual calculations: continuity hypotheses in later results must be checked exactly as stated, rather than inferred from a partial test.
Check Your Understanding
For each question, identify which part of the continuity test is relevant.
- A function is defined at \(c\), and its limit at \(c\) is \(3\), but \(f(c)=7\). Is it continuous at \(c\)? Explain.
- The formula for \(q(x)\) has denominator \(x-4\). Can \(q\) be continuous at \(4\) if its domain excludes \(4\), even if a finite limit exists there?
- At an interior point, the right-hand limit equals the function value, but the left-hand limit differs. Is the function continuous there?
- At the left endpoint of a domain \([a,\infty)\), which one-sided limit is used to test continuity relative to that domain?
- State the three checks required for continuity at an interior point of an interval.
