Why Continuity Must Be Checked Point by Point
The Extreme Value Theorem guarantees that a function attains both an absolute maximum and an absolute minimum if it is continuous on a closed, bounded interval. Checking that the interval has the required form is only part of the task: the function must also be continuous at every point of that interval.
For a formula that is continuous throughout its domain, this check may be brief. For a piecewise-defined function, or a function whose formula has a denominator or other restriction, the important points are often where the formula changes, where it is undefined, and at the interval’s endpoints.
Continuity at Interior Points and Endpoints
At an interior point \(c\), continuity requires the function’s limit as \(x\) approaches \(c\) to exist and equal the function’s value there. Equivalently, both one-sided limits must exist and agree with \(f(c)\).
The directions at the endpoints matter. A function defined only on \([a,b]\) need not have values to the left of \(a\) or to the right of \(b\). Continuity on the interval therefore uses only the limit from within the interval at each endpoint.
For a piecewise-defined function, the same definition applies at every point. At an interior point where the rule changes, compare the limit from the left, the limit from the right, and the function’s actual value at that point. These three quantities must all agree.
Why the One-Sided Tests Work
At an interior point \(c\), the definition of continuity requires \(f(x)\) to approach \(f(c)\) as \(x\) approaches from either direction. In limit notation, this means the left-hand limit and right-hand limit must both exist and both equal \(f(c)\).
For completeness, suppose \(\lim_{x\to c^-}f(x)=f(c)\) and \(\lim_{x\to c^+}f(x)=f(c)\). Given any \(\varepsilon>0\), the left-hand limit provides a distance \(\delta_1>0\) such that \(0<c-x<\delta_1\) implies \(|f(x)-f(c)|<\varepsilon\). The right-hand limit provides \(\delta_2>0\) such that \(0<x-c<\delta_2\) implies the same inequality. Choose \(\delta=\min(\delta_1,\delta_2)\). Then whenever \(0<|x-c|<\delta\), \(x\) is on one side or the other, and in either case \(|f(x)-f(c)|<\varepsilon\). Thus \(f\) is continuous at \(c\). Conversely, if \(f\) is continuous at \(c\), its defining limit condition holds from each side, so both one-sided limits equal \(f(c)\).
At an endpoint, only one direction lies inside the interval, so only that one-sided limit is required. This is why endpoint checks are part of continuity on a closed interval, even though they are not two-sided checks.
Continuity Facts That Make the Check Efficient
A continuity check does not require a fresh limit calculation at every point if a standard continuity fact applies. Polynomials are continuous everywhere. A quotient of continuous functions is continuous wherever its denominator is nonzero. Sums, products, and compositions of continuous functions are continuous wherever they are defined.
These facts establish continuity within regions where a formula is valid. They do not automatically settle what happens at a formula change, a point excluded by a denominator, or an endpoint. Those points must still be checked against the definition.
Include both endpoints, every piecewise boundary, and any point where a formula is undefined or its continuity rule does not apply.
Use known continuity facts for the formula on that region.
Compare the appropriate one-sided limits with one another and with the function value.
The function is continuous on the closed interval only if every required point passes its check.
Worked Examples
Worked Example: A Quotient Rule at the Left Endpoint
Define \(f\) on \([1,3]\) by \(f(1)=2\) and \(f(x)=\frac{x^2-1}{x-1}\) for \(1<x\leq 3\). Determine whether \(f\) is continuous on \([1,3]\).
For \(1<x\leq 3\), factor the numerator:
Since \(x-1\ne 0\) for \(x>1\), cancellation is valid throughout that part of the interval. The function is therefore continuous at each point with \(1<x\leq 3\), using the polynomial expression \(x+1\) there.
At the left endpoint, use the limit from within \([1,3]\):
Thus \(f\) is continuous at \(1\) as well, and hence continuous on \([1,3]\). The endpoint value agrees with the limiting value, which is essential; the original quotient alone is not defined at \(x=1\).
Check: For any \(x>1\), the simplified expression gives \(x+1\), and at \(x=1\) the defined value is \(2=1+1\). The endpoint value matches the expression’s limiting value.
Worked Example: Checking a Piecewise Boundary and Both Endpoints
On \([-2,2]\), let \(f(x)=x^2\) for \(x\leq 0\), and \(f(x)=2x\) for \(x>0\). Determine whether \(f\) is continuous on the interval.
Within \([-2,0]\), the rule \(x^2\) is a polynomial, so it is continuous there. Within \((0,2]\), the rule \(2x\) is also a polynomial, so it is continuous there. The point \(0\) is the boundary that must be checked separately. Its function value comes from the first branch:
The left-hand limit uses the first rule, and the right-hand limit uses the second:
Both one-sided limits equal \(f(0)=0\), so \(f\) is continuous at the boundary. At the endpoints, the one-sided checks are
All points within each branch are covered by polynomial continuity, the boundary at \(0\) passes the two-sided check, and both endpoints pass their one-sided checks. Therefore \(f\) is continuous on \([-2,2]\).
Check: The left and right limits at \(0\) are both \(0\), matching the defined value; the endpoint limits are \(4\), matching the corresponding endpoint values.
Worked Example: A Piecewise Function That Fails at Its Boundary
On \([-2,2]\), define \(g(x)=x+1\) for \(-2\leq x<0\), \(g(0)=3\), and \(g(x)=x+2\) for \(0<x\leq 2\). Determine whether \(g\) is continuous on \([-2,2]\).
Each formula is a polynomial on the part of the interval where it applies, so check the interior boundary \(x=0\) separately. From the left and right, the limits are
Because \(1\ne 2\), the two-sided limit at \(0\) does not exist. Also, the defined value \(g(0)=3\) agrees with neither one-sided limit. Thus \(g\) is not continuous at \(0\), so it is not continuous on \([-2,2]\).
The endpoint checks do not repair this failure. They do pass:
Check: The boundary limits are \(1\) and \(2\), which are unequal, and the boundary value is \(3\). This verifies the specific point where continuity fails.
From the Check to the Extreme Value Theorem
For the Extreme Value Theorem, the relevant conditions are that the function is continuous on the entire interval and that the interval is closed and bounded. If both conditions hold, the theorem guarantees that the function attains an absolute maximum and an absolute minimum somewhere on the interval.
For example, the function in the second worked example is continuous on the closed, bounded interval \([-2,2]\). Its continuity check, together with the interval’s form, establishes that the theorem’s hypotheses are met. This conclusion is about what the theorem guarantees; it does not locate the maximum or minimum.
If even one required point fails the continuity test, the function is not continuous on the whole interval, and the theorem’s continuity hypothesis has not been established. That is a precise statement about whether the theorem can be invoked; it does not, by itself, determine whether the function has absolute extrema.
Common Pitfall: Checking Only the Formula
A common error is to observe that each branch is a polynomial and conclude that the whole piecewise function is continuous. For the function \(g\) above, both branch formulas are polynomials, but their limits at \(0\) are \(1\) and \(2\). Continuity of each formula on its own region does not guarantee continuity where the rules meet.
Another error is to use a two-sided limit at an endpoint. For \([a,b]\), the interval includes no points to the left of \(a\) or to the right of \(b\). The correct tests are \(\lim_{x\to a^+}f(x)=f(a)\) and \(\lim_{x\to b^-}f(x)=f(b)\).
Check Your Understanding
For each question, identify which points require special attention and state what must be compared.
- What one-sided limit must be compared with \(f(a)\) to check continuity at the left endpoint of \([a,b]\)?
- A piecewise function has polynomial rules on either side of an interior boundary \(c\). What three quantities must agree for continuity at \(c\)?
- Suppose a quotient has a denominator that is nonzero everywhere on \([a,b]\). What continuity fact applies within the interval, and which points should still be checked explicitly?
- For the piecewise function in the second worked example, what are the left-hand limit, right-hand limit, and function value at \(0\)?
- If a function fails continuity at one point of a closed interval, what can you conclude about using the Extreme Value Theorem’s continuity hypothesis?
