What Does It Mean for a Function to Be Continuous at a Point?
A function can behave smoothly near an input without being defined at that input. It can also have a value at the input while its nearby values fail to approach a single number. To decide whether the function is continuous at a point, we must check three separate facts.
Let \(f\) be defined for inputs in some interval around \(a\), possibly with the value at \(a\) missing. Continuity at \(a\) requires that \(f(a)\) exists, that the finite two-sided limit as \(x\) approaches \(a\) exists, and that this limit equals \(f(a)\). Checking the conditions separately makes it clear why a function may fail to be continuous.
- \(f(a)\) is defined.
- \(\lim_{x\to a} f(x)\) exists as a finite real number.
- \(\lim_{x\to a} f(x)=f(a)\).
The limit concerns values of \(f(x)\) for inputs near \(a\), not the value at \(a\) itself. In the limit process, \(x\ne a\). The function value at \(a\) and the limit at \(a\) are therefore distinct pieces of information; continuity requires them to match.
The Three Conditions, One at a Time
First, the function value must exist. The expression defining \(f\) must assign a value to \(a\). For instance, if \(f(x)=x+1\) only for \(x\ne 2\), then \(f(2)\) is not defined. Even if the nearby values have a limit as \(x\) approaches \(2\), the first continuity condition fails.
Second, the two-sided limit must exist and be finite. Values from both sides of \(a\) must approach the same real number. If the left-hand and right-hand limits differ, there is no two-sided limit. If the function grows without bound or oscillates without settling to one value, there is no finite two-sided limit either.
In particular, when both one-sided limits are finite, the two-sided limit exists exactly when the left-hand and right-hand limits exist and are equal. This provides a useful way to test the second condition.
Third, the limit must equal the function value. A function may have a defined value and a finite limit, yet those numbers may differ. In that case the nearby values approach one number while the function assigns another value at \(a\), so continuity fails.
Is \(f(a)\) defined? If not, continuity fails.
Does \(\lim_{x\to a}f(x)\) exist as a finite real number? If not, continuity fails.
If both exist, is \(\lim_{x\to a}f(x)=f(a)\)? If so, \(f\) is continuous at \(a\).
Worked Examples
Worked Example: A Function That Is Continuous at the Point
Test \(f(x)=x^2+3x\) for continuity at \(a=2\).
Condition 1: Evaluate the function.
So \(f(2)\) is defined. The first condition holds.
Condition 2: Find the limit. For \(x\ne 2\), rewrite the difference from the candidate value:
As \(x\) approaches \(2\), \(x-2\) approaches \(0\) and \(x+5\) approaches \(7\). Thus \(f(x)-10\) approaches \(0\), so \(\lim_{x\to 2}f(x)=10\).
Condition 3: Compare.
All three conditions hold, so \(f\) is continuous at \(2\). As a check, the difference \(f(x)-f(2)\) is exactly \((x-2)(x+5)\), which tends to zero as \(x\) approaches \(2\).
Worked Example: The Function Value and Limit Do Not Match
Let \(f(x)=\dfrac{x^2-1}{x-1}\) when \(x\ne 1\), and let \(f(1)=4\). Test continuity at \(a=1\).
Condition 1: The definition assigns \(f(1)=4\), so the function value exists.
Condition 2: For \(x\ne 1\), factor the numerator and cancel the common factor:
This cancellation is valid for the nearby inputs \(x\ne 1\) used in the limit. Therefore,
The finite two-sided limit exists. But \(2\ne 4\), since \(f(1)=4\). The third condition fails, so \(f\) is not continuous at \(1\). Checking by substitution into the original definition confirms the distinction: the assigned value at \(1\) is \(4\), whereas the nearby formula \(x+1\) approaches \(2\).
Worked Example: A Limit Exists but the Function Value Does Not
Suppose \(f(x)=x+1\) for \(x\ne 2\), and \(f(2)\) is not defined. Test continuity at \(a=2\).
Condition 1: There is no assigned value \(f(2)\), so the first condition fails.
Condition 2: For nearby inputs other than \(2\), the formula is \(x+1\). Hence
This is a finite two-sided limit. However, it cannot equal \(f(2)\), because \(f(2)\) is not defined. The function is not continuous at \(2\). Substituting any nearby input into the formula gives \(x+1\), whose values approach \(3\); this checks the limit calculation but does not supply the missing function value.
Worked Example: The Two-Sided Limit Does Not Exist
Define \(f(0)=8\) and \(f(x)=\dfrac{x}{|x|}\) for \(x\ne 0\). Test continuity at \(a=0\).
Condition 1: The function value exists and is \(f(0)=8\).
Condition 2: If \(x<0\), then \(|x|=-x\), so \(x/|x|=-1\). If \(x>0\), then \(|x|=x\), so \(x/|x|=1\). Thus
The one-sided limits are unequal, so the two-sided limit does not exist. The second condition fails, and \(f\) is not continuous at \(0\). For a direct check, values from the left are always \(-1\), while values from the right are always \(1\); they cannot approach one common number.
Why the Three Conditions Are Equivalent to Continuity
The three-part test is not just a convenient checklist. It states precisely what it means for nearby function values to approach the value assigned at the point.
For a function defined at \(a\), the epsilon-delta definition of continuity says that for every \(\varepsilon>0\), there is a \(\delta>0\) such that whenever \(x\) is in the domain and \(|x-a|<\delta\), then \(|f(x)-f(a)|<\varepsilon\). This definition already requires \(f(a)\) to exist. The inequality says that inputs sufficiently close to \(a\) produce outputs as close as desired to \(f(a)\).
For a two-sided limit, the corresponding definition uses \(0<|x-a|<\delta\): the strict inequality excludes \(x=a\). A finite limit \(L\) means that for every \(\varepsilon>0\), some \(\delta>0\) ensures \(|f(x)-L|<\varepsilon\) whenever \(0<|x-a|<\delta\).
To see why, first suppose \(f\) is continuous at \(a\). Given any \(\varepsilon>0\), continuity supplies a \(\delta>0\) such that \(|f(x)-f(a)|<\varepsilon\) whenever \(0<|x-a|<\delta\). This is exactly the limit condition with \(L=f(a)\). Therefore the limit exists and equals \(f(a)\).
Conversely, suppose \(f(a)\) is defined and \(\lim_{x\to a}f(x)=f(a)\). By the definition of that limit, for every \(\varepsilon>0\) there is a \(\delta>0\) such that \(|f(x)-f(a)|<\varepsilon\) whenever \(0<|x-a|<\delta\). At \(x=a\), the difference is \(|f(a)-f(a)|=0<\varepsilon\). So the same closeness condition also holds when \(x=a\), and hence for every \(x\) with \(|x-a|<\delta\). This is the epsilon-delta definition of continuity. The two formulations agree.
A Common Mistake: Substituting Without Checking the Point
A frequent error is to simplify or substitute in a formula, find a limit, and then conclude continuity without checking the function's assigned value. The limit describes nearby outputs; it does not tell us what value, if any, the function has at \(a\).
For example, in the second worked example, canceling \(x-1\) gives the nearby expression \(x+1\), whose limit is \(2\). The incorrect conclusion would be: “The limit is \(2\), so the function is continuous at \(1\).” But the definition assigns \(f(1)=4\), and \(2\ne 4\). The correct conclusion is that the limit exists but does not equal the function value, so continuity fails.
Check Your Understanding
For each question, identify which of the three continuity conditions are satisfied and state whether the function is continuous at the specified point.
- Let \(f(x)=x^2-1\). At \(a=2\), find \(f(2)\) and \(\lim_{x\to2}f(x)\).
- Let \(g(x)=\dfrac{x^2-4}{x-2}\) for \(x\ne2\), and let \(g(2)=7\). Does \(g\) satisfy all three conditions at \(2\)?
- Let \(h(x)=x-3\) for \(x\ne3\), with no value assigned at \(3\). Which condition fails at \(3\), and what is the limit there?
- Suppose \(p(0)=5\), \(\lim_{x\to0^-}p(x)=2\), and \(\lim_{x\to0^+}p(x)=2\). Is \(p\) continuous at \(0\)? Explain.
- Suppose \(q(1)=4\), but the left-hand and right-hand limits at \(1\) are \(3\) and \(6\). Which continuity condition fails?
