Diagnosing Continuity from Different Representations
A graph can show a hole or a jump, a table can reveal different behavior on the two sides of an input, and a formula can allow an exact limit calculation. These are different kinds of evidence for the same question: is the function continuous at the specified input?
A reliable diagnosis checks the function value and the behavior of nearby outputs separately. A graph or a table may suggest what is happening, but when possible, use the formula or the stated graph behavior to justify the limit. In particular, a handful of table entries cannot by itself prove what happens arbitrarily close to an input.
These are three separate checks. The function value is the output assigned exactly at \(a\). The limit describes the output that \(f(x)\) approaches as \(x\) approaches \(a\), without requiring \(x=a\). A function is continuous at \(a\) only when both are defined in the required sense and agree.
For an interior input, the two-sided limit exists as a finite number precisely when the left-hand and right-hand limits both exist as finite numbers and are equal. Thus, when reading a graph or table, compare what happens from each side; the value at the input is a separate check.
What Each Representation Can Tell You
On a graph, a filled point at \((a,y)\) represents the assigned function value \(f(a)=y\). An open circle marks a point not included at that location, but the graph’s nearby branches may still approach its height. The branches indicate the limiting behavior; the filled point indicates the function value. They need not agree.
A table lists function values at selected inputs. Values on both sides of \(a\) can suggest whether outputs approach the same number, different numbers, or grow without bound. However, any finite table samples only finitely many inputs. Unless additional information, such as a formula or an explicit description of the graph, supports the conclusion, the table alone does not establish the limit.
A formula can often settle the issue exactly. Direct substitution works when the formula is defined and its component functions are continuous at the input. If substitution produces an indeterminate expression, algebraic simplification may reveal the limit, but the simplified expression for nearby inputs does not automatically redefine the original function at \(a\).
Worked Example: A Graph with a Hole and a Different Function Value
Suppose a graph shows two branches approaching the open circle \((2,3)\) from the left and right. The graph also has a filled point at \((2,1)\). Diagnose continuity at \(x=2\).
The filled point gives the function value:
The two branches approach height \(3\) from both sides. Therefore the one-sided limits agree, so the two-sided limit exists and is
Now check the three conditions. First, \(f(2)=1\) is defined. Second, the finite limit exists and equals \(3\). Third, the limit is not equal to the function value, since \(3\ne 1\). The function is therefore discontinuous at \(2\), specifically because the third condition fails. The graph’s open circle indicates the limiting height, not the assigned value.
As a final check, compare the two relevant quantities directly: the filled point has height \(1\), while both branches approach height \(3\). Since these do not match, continuity fails.
Using a Table Without Overclaiming
For a two-sided limit, organize nearby inputs into a left-hand group and a right-hand group. Do not use the row \(x=a\) to determine the limit: that row gives \(f(a)\), not what nearby function values approach. Look for a consistent trend on each side, then compare the two sides.
Worked Example: A Table Suggesting a Jump
Suppose the table and the accompanying graph description give the following information. The graph’s left branch approaches \(2\) as \(x\) approaches \(1\) from the left, and its right branch approaches \(-1\) as \(x\) approaches \(1\) from the right. The table is consistent with those branches:
| \(x\) | \(f(x)\) |
|---|---|
| \(0.90\) | \(2.01\) |
| \(0.99\) | \(2.0001\) |
| \(1.01\) | \(-0.99\) |
| \(1.10\) | \(-0.90\) |
The entries with \(x<1\) are close to \(2\), while those with \(x>1\) are close to \(-1\). The stated graph behavior confirms the one-sided limits:
Because \(2\ne-1\), the two-sided limit does not exist. Whatever value the function may assign at \(1\), condition 2 fails, so the function is discontinuous there. The table supports the diagnosis, but the exact conclusion relies on the supplied graph behavior rather than on four sampled values alone.
Check the logic: a two-sided limit would require the left- and right-hand limits to be the same finite number. Here their exact stated values differ, so no choice of \(f(1)\) can make the function continuous at \(1\).
Testing a Formula at a Point
When a formula is available, first distinguish its value at the input from its values nearby. A factor that cancels during limit computation can simplify the nearby behavior, but the original function’s domain and assigned point value still determine condition 1 and condition 3.
Worked Example: Simplifying a Formula and Checking the Defined Value
Let
Determine whether \(f\) is continuous at \(x=2\). The piecewise definition directly gives
For \(x\ne 2\), factor the numerator and simplify:
This cancellation is valid for \(x\ne 2\), which is exactly the restriction on this branch of the formula. For inputs approaching \(2\) but not equal to \(2\), the function therefore has the same values as \(x+2\), and
The function is defined at \(2\), the finite limit exists, and the limit equals the function value: \(4=f(2)\). Thus \(f\) is continuous at \(2\).
As a check on the limit calculation, the simplified nearby expression \(x+2\) approaches \(4\) as \(x\) approaches \(2\). This matches the separately specified value \(f(2)=4\).
Why the Three Conditions Are Necessary
The definition of a limit says that \(\lim_{x\to a}f(x)=L\) when outputs can be made arbitrarily close to \(L\) by taking inputs sufficiently close to, but different from, \(a\). In precise terms, for every \(\varepsilon>0\), there must be a \(\delta>0\) such that
Continuity means that this limiting output is the actual output at \(a\). Setting \(L=f(a)\) in the limit condition gives the corresponding continuity test:
At \(x=a\), the output in this last inequality is exactly \(f(a)\), so its distance from \(f(a)\) is zero. For \(x\ne a\), the statement is precisely the limit condition with \(L=f(a)\). This explains why continuity requires both a defined function value and a finite limit equal to it: neither nearby behavior alone nor the point value alone is enough.
Diagnosing Other Failures
A graph that rises or falls without bound near \(a\), or a formula that has a vertical asymptote there, indicates that no finite limit exists. The function may or may not be assigned a value at \(a\); either way, condition 2 fails. A missing point in the domain means condition 1 fails. If the finite limit exists but the assigned value differs from it, condition 3 fails.
Worked Example: A Defined Value with No Finite Limit
Define
The function value is defined: \(f(3)=7\). For inputs on either side of \(3\), the squared denominator is positive and approaches \(0\). For example,
and at distance \(0.01\),
As \(x\) approaches \(3\) from either side, the denominator approaches \(0\) through positive values, so \(f(x)\) grows without bound. There is no finite two-sided limit. Condition 1 holds, condition 2 fails, and condition 3 cannot hold because there is no finite limit to equal \(f(3)\). Therefore \(f\) is discontinuous at \(3\).
The increasing values provide a numerical check of the behavior, while the formula explains why the values grow without bound as the distance to \(3\) decreases.
A Practical Diagnosis Routine
Use the filled point on a graph, the \(x=a\) row in a table, or the formula’s definition at \(a\). Record whether \(f(a)\) exists.
Use the graph’s branches, table trends, or one-sided formula limits. Determine whether both sides approach the same finite number.
If a finite two-sided limit exists, compare it with \(f(a)\). Continuity requires exact equality.
State whether the value is undefined, the finite limit does not exist, or the limit and value are unequal. If none fails, the function is continuous at \(a\).
Common Mistake: Treating the Limit as the Function Value
A common error is to cancel a factor in a formula and then use the simplified expression to claim that the original function has that value at the excluded input. For example, if
then the incorrect calculation is
The cancellation is valid only when \(x\ne 2\). The original formula does not define \(g(2)\), so condition 1 fails; the formula alone does not make \(g\) continuous at \(2\). The correct conclusion is
A separate definition assigning \(g(2)=4\) would make the limit and value agree, as in the earlier example. Always report the limit and the function value separately before deciding continuity.
Check Your Understanding
For each situation, identify whether the function is continuous at the specified input and name any failed condition.
- A graph’s two branches approach height \(5\) at \(x=0\), and its filled point at \(x=0\) is \((0,5)\). Is the function continuous there?
- A graph approaches height \(2\) from the left and height \(4\) from the right at \(x=3\). Which continuity condition fails?
- A table lists \(f(a)=6\) and several nearby outputs close to \(6\). Does this finite table alone prove continuity at \(a\)? Explain.
- For \(h(x)=(x^2-9)/(x-3)\) when \(x\ne3\), with \(h(3)=6\), compute the limit at \(3\) and decide continuity.
- A function is defined at \(a\), but its values grow without bound as \(x\) approaches \(a\) from both sides. Which condition fails?
