The Central Idea
A function \(f\) is continuous at \(x=a\) when the value of the function agrees with the value approached by the function nearby.
Symbolically,
\[ \boxed{\lim_{x\to a}f(x)=f(a)}. \]This compact equation summarizes the entire idea of continuity at a point.
The Three Conditions for Continuity at a Point
A function \(f\) is continuous at \(x=a\) when all three conditions hold:
- \(f(a)\) is defined.
- \(\lim_{x\to a}f(x)\) exists.
- \(\lim_{x\to a}f(x)=f(a)\).
If even one of these conditions fails, the function is not continuous at \(x=a\).
Condition 1: The Function Must Be Defined
To be continuous at \(x=a\), the function must have an actual value there.
That means
\[ f(a) \]must exist.
A hole in the graph immediately causes this first condition to fail.
Condition 2: The Limit Must Exist
The nearby values of the function must approach one common finite value:
\[ \lim_{x\to a}f(x). \]If the left-hand and right-hand limits disagree, the two-sided limit does not exist, so the function cannot be continuous at that point.
Condition 3: The Limit Must Equal the Function Value
Even if \(f(a)\) exists and the limit exists, continuity still requires
\[ \boxed{\lim_{x\to a}f(x)=f(a)}. \]If the nearby values approach one number but the function is assigned a different value at \(x=a\), the function is discontinuous there.
A Simple Continuous Example
Consider
\[ f(x)=x^2+1. \]At \(x=3\),
\[ f(3)=10. \]Also,
\[ \lim_{x\to3}(x^2+1)=10. \]Therefore,
\[ \boxed{\lim_{x\to3}f(x)=f(3)} \]and \(f\) is continuous at \(x=3\).
Polynomials Are Continuous Everywhere
Every polynomial is continuous for every real input.
For a polynomial \(P(x)\),
\[ \boxed{\lim_{x\to a}P(x)=P(a)} \]for every real number \(a\).
This is why direct substitution works so reliably for polynomial limits.
Rational Functions Are Continuous on Their Domains
A rational function
\[ R(x)=\frac{P(x)}{Q(x)} \]is continuous wherever
\[ Q(x)\neq0. \]Any input that makes the denominator zero must be examined separately.
Example: Rational Continuity
Consider
\[ f(x)=\frac{x+1}{x-2}. \]This function is continuous for every real \(x\) except
\[ x=2. \]At every other input, direct substitution gives the function value and the limit.
A Hole Breaks Continuity
Consider
\[ f(x)=\frac{x^2-4}{x-2}. \]For \(x\neq2\),
\[ f(x)=x+2. \]The limit exists:
\[ \lim_{x\to2}f(x)=4. \]But the original function is undefined at \(x=2\).
Therefore, \(f\) is not continuous at \(x=2\).
A Function Value Can Break Continuity
Suppose
\[ f(x)= \begin{cases} x+2, & x\neq2,\\ 10, & x=2. \end{cases} \]Then
\[ \lim_{x\to2}f(x)=4, \]but
\[ f(2)=10. \]Since the values do not match,
\[ \boxed{\lim_{x\to2}f(x)\neq f(2)}, \]so the function is discontinuous at \(x=2\).
Continuity and Graphs
Informally, a graph is continuous at a point when it passes through that point without a hole, jump, or vertical break.
The common phrase “draw it without lifting your pencil” can build intuition, but the formal definition is still
\[ \lim_{x\to a}f(x)=f(a). \]The Pencil Test Has Limits
The pencil description is helpful but not mathematically complete.
A graph may be difficult to draw accurately, and continuity can be studied even when no graph is available.
The limit-based definition is the reliable standard.
Continuity from the Left
At the right endpoint of an interval, continuity is naturally tested from the left.
A function is continuous from the left at \(x=b\) when
\[ \boxed{\lim_{x\to b^-}f(x)=f(b)}. \]Continuity from the Right
At the left endpoint of an interval, continuity is naturally tested from the right.
A function is continuous from the right at \(x=a\) when
\[ \boxed{\lim_{x\to a^+}f(x)=f(a)}. \]Continuity on an Open Interval
A function is continuous on an open interval
\[ (a,b) \]if it is continuous at every point inside that interval.
That means
\[ \lim_{x\to c}f(x)=f(c) \]for every
\[ c\in(a,b). \]Continuity on a Closed Interval
A function is continuous on
\[ [a,b] \]when:
- it is continuous at every point in \((a,b)\);
- it is continuous from the right at \(a\);
- it is continuous from the left at \(b\).
Why Endpoints Use One-Sided Limits
At the left endpoint \(a\) of
\[ [a,b], \]the interval contains no points immediately to the left.
So only
\[ \lim_{x\to a^+}f(x) \]is relevant.
The same reasoning applies from the left at \(b\).
Example: Continuity on a Closed Interval
Consider
\[ f(x)=\sqrt{x} \]on
\[ [0,9]. \]The function is continuous at every interior point in
\[ (0,9). \]At \(0\),
\[ \lim_{x\to0^+}\sqrt{x}=0=f(0). \]At \(9\),
\[ \lim_{x\to9^-}\sqrt{x}=3=f(9). \]Therefore, \(\sqrt{x}\) is continuous on \([0,9]\).
Standard Continuous Function Families
Many familiar function families are continuous throughout their natural domains.
- polynomial functions;
- rational functions where the denominator is nonzero;
- root functions where defined;
- exponential functions;
- logarithmic functions on positive inputs;
- sine and cosine functions.
Continuity Is Preserved by Addition
If \(f\) and \(g\) are continuous at \(x=a\), then
\[ f+g \]is also continuous at \(a\).
Symbolically,
\[ \lim_{x\to a}[f(x)+g(x)] = f(a)+g(a). \]Continuity Is Preserved by Subtraction and Multiplication
If \(f\) and \(g\) are continuous at \(a\), then so are
\[ f-g \]and
\[ fg. \]These closure properties allow complicated continuous functions to be built from simpler ones.
Continuity Is Preserved by Quotients When Defined
If \(f\) and \(g\) are continuous at \(a\) and
\[ g(a)\neq0, \]then
\[ \frac{f}{g} \]is continuous at \(a\).
Continuity Is Preserved by Composition
Suppose \(g\) is continuous at \(a\), and \(f\) is continuous at
\[ g(a). \]Then the composition
\[ f(g(x)) \]is continuous at \(a\).
This principle explains why many complicated formulas remain continuous on their domains.
Direct Substitution and Continuity
When a function is continuous at \(x=a\), evaluating the limit is simple:
\[ \boxed{\lim_{x\to a}f(x)=f(a)}. \]This is why direct substitution is so powerful for continuous functions.
Example: Composite Continuity
Evaluate
\[ \lim_{x\to2}\sqrt{x^2+5}. \]The polynomial
\[ x^2+5 \]is continuous, and the square-root function is continuous for positive inputs.
Therefore,
\[ \lim_{x\to2}\sqrt{x^2+5} = \sqrt{2^2+5} = \boxed{3}. \]Piecewise Functions Require Boundary Checks
A piecewise function may be continuous on each piece but fail to be continuous where the formulas change.
Those boundary points must be checked separately.
Example: Continuous Piecewise Function
Consider
\[ f(x)= \begin{cases} x+1, & x<2,\\ 5-x, & x\ge2. \end{cases} \]From the left,
\[ \lim_{x\to2^-}f(x)=3. \]From the right,
\[ \lim_{x\to2^+}f(x)=3. \]Also,
\[ f(2)=3. \]Therefore, \(f\) is continuous at \(x=2\).
Example: Discontinuous Piecewise Function
Consider
\[ g(x)= \begin{cases} x+1, & x<2,\\ x+4, & x\ge2. \end{cases} \]Then
\[ \lim_{x\to2^-}g(x)=3 \]while
\[ \lim_{x\to2^+}g(x)=6. \]The two-sided limit does not exist, so \(g\) is not continuous at \(x=2\).
Choosing a Parameter for Continuity
Piecewise functions often contain an unknown parameter that can be chosen to make the function continuous.
Suppose
\[ f(x)= \begin{cases} x+3, & x<2,\\ k, & x=2,\\ 7-x, & x>2. \end{cases} \]The left-hand and right-hand limits are both
\[ 5. \]Therefore, continuity requires
\[ \boxed{k=5}. \]Continuity Can Be Repaired at a Hole
If a function has a removable hole at \(x=a\) but
\[ \lim_{x\to a}f(x)=L, \]we can define
\[ f(a)=L \]to create a continuous extension.
Example: Continuous Extension
Let
\[ f(x)=\frac{x^2-9}{x-3} \]for \(x\neq3\).
Since
\[ \frac{x^2-9}{x-3}=x+3 \qquad(x\neq3), \]we have
\[ \lim_{x\to3}f(x)=6. \]Defining
\[ \boxed{f(3)=6} \]makes the extended function continuous at \(3\).
Continuity Is a Local Property
To determine whether a function is continuous at \(x=a\), only nearby behavior matters.
The function may behave very differently far away without affecting continuity at \(a\).
Continuity on an Interval Is a Global Statement Built from Local Ones
Saying a function is continuous on an interval means it passes the local continuity test at every relevant point in that interval.
So interval continuity is assembled from point-by-point continuity.
Continuity and Numerical Tables
Numerically, continuity near \(x=a\) means nearby outputs approach the same value as \(f(a)\).
A table should show:
- inputs approaching \(a\) from both sides;
- outputs approaching one common value;
- that common value matching the actual function value.
Continuity and Graphs
Graphically, continuity at \(x=a\) means:
- the graph approaches one height from the left;
- the graph approaches the same height from the right;
- the graph actually includes the point at that height.
Continuity and Symbolic Formulas
Symbolically, continuity is often verified by:
- finding \(f(a)\);
- finding \(\lim_{x\to a}f(x)\);
- checking whether the two values are equal.
A Reliable Procedure at a Point
- Compute \(f(a)\).
- Check the left-hand limit.
- Check the right-hand limit.
- Confirm the two-sided limit exists.
- Compare the limit with \(f(a)\).
- If they match, the function is continuous at \(a\).
A Reliable Procedure on an Interval
- Identify the interval and its endpoints.
- Check continuity throughout the interior.
- Locate any points where formulas change or denominators vanish.
- Check those points separately.
- At a closed left endpoint, use right continuity.
- At a closed right endpoint, use left continuity.
- State the full interval on which continuity holds.
Example: Full Point-Continuity Workflow
Determine whether
\[ f(x)= \begin{cases} x^2, & x\neq2,\\ 4, & x=2 \end{cases} \]is continuous at \(x=2\).
First,
\[ f(2)=4. \]Next,
\[ \lim_{x\to2}x^2=4. \]Therefore,
\[ \boxed{\lim_{x\to2}f(x)=f(2)=4}. \]The function is continuous at \(x=2\).
Example: Continuity Failure by a Jump
Suppose
\[ f(x)= \begin{cases} 2x, & x<1,\\ x+3, & x\ge1. \end{cases} \]Then
\[ \lim_{x\to1^-}f(x)=2 \]while
\[ \lim_{x\to1^+}f(x)=4. \]The two-sided limit does not exist, so the function is not continuous at \(x=1\).
Example: Find the Interval of Continuity
Consider
\[ f(x)=\frac{1}{x^2-9}. \]The denominator factors as
\[ x^2-9=(x-3)(x+3). \]The function is undefined at
\[ x=-3 \qquad\text{and}\qquad x=3. \]Therefore, \(f\) is continuous on
\[ \boxed{(-\infty,-3)\cup(-3,3)\cup(3,\infty)}. \]Common Error: Checking Only the Function Value
Knowing that \(f(a)\) exists does not prove continuity.
The nearby limit must also exist and equal that value.
Common Error: Checking Only the Limit
A limit can exist even when the function is undefined at the target.
Therefore,
\[ \lim_{x\to a}f(x) \]existing does not by itself guarantee continuity.
Common Error: Ignoring the Endpoint Direction
At the left endpoint of a closed interval, use a right-hand limit.
At the right endpoint, use a left-hand limit.
A two-sided limit may not even be meaningful within the restricted domain.
Common Error: Assuming Every Formula Is Continuous Everywhere
A rational formula may fail where its denominator is zero.
A square-root formula may have a restricted real domain.
A logarithm requires positive input.
Continuity must always be interpreted relative to the function's domain.
Common Error: Confusing Continuity with Smoothness
A function can be continuous even if it has a sharp corner or cusp.
Continuity requires no break in the function value.
Smoothness involves stronger conditions connected to derivatives.
Practice 1: Polynomial Continuity
Determine whether
\[ f(x)=x^3-2x+5 \]is continuous at \(x=2\).
Practice 2: Rational Continuity
Determine whether
\[ f(x)=\frac{x+1}{x-4} \]is continuous at \(x=3\).
Practice 3: Undefined Point
Is
\[ f(x)=\frac{x^2-1}{x-1} \]continuous at \(x=1\)?
Practice 4: Repairing a Hole
Define
\[ f(x)=\frac{x^2-1}{x-1} \]for \(x\neq1\).
What value should be assigned to \(f(1)\) to make the function continuous?
Practice 5: Piecewise Continuity
Let
\[ f(x)= \begin{cases} x+2, & x<3,\\ 8-x, & x\ge3. \end{cases} \]Is \(f\) continuous at \(x=3\)?
Practice 6: Parameter for Continuity
Let
\[ f(x)= \begin{cases} 2x+1, & x<2,\\ k, & x=2,\\ x+3, & x>2. \end{cases} \]Find \(k\) so that \(f\) is continuous at \(x=2\).
Practice 7: Closed-Interval Endpoint
Explain why
\[ f(x)=\sqrt{x} \]is continuous at the left endpoint \(x=0\) of its domain.
Practice 8: Interval of Continuity
Find the intervals on which
\[ f(x)=\frac{1}{x^2-4} \]is continuous.
Practice 9: Limit Exists but Continuity Fails
Suppose
\[ \lim_{x\to5}f(x)=7 \]but
\[ f(5)=10. \]Is \(f\) continuous at \(x=5\)?
Practice 10: State the Three Conditions
State the three conditions required for \(f\) to be continuous at \(x=a\).
Practice Answers
Practice 1
Polynomials are continuous everywhere.
\[ f(2)=2^3-2(2)+5=9. \]Therefore,
\[ \boxed{\lim_{x\to2}f(x)=f(2)=9}. \]Practice 2
The denominator is nonzero at \(x=3\):
\[ 3-4=-1. \]Therefore, the rational function is continuous at \(3\).
\[ \boxed{f(3)=-4}. \]Practice 3
The original function is undefined at \(x=1\), so it is not continuous there.
However,
\[ \lim_{x\to1}\frac{x^2-1}{x-1}=2. \]Practice 4
Since
\[ \frac{x^2-1}{x-1}=x+1 \qquad(x\neq1), \]the limit at \(1\) is \(2\).
Define
\[ \boxed{f(1)=2}. \]Practice 5
From the left,
\[ \lim_{x\to3^-}f(x)=5. \]From the right,
\[ \lim_{x\to3^+}f(x)=5. \]Also,
\[ f(3)=5. \]Therefore, \(f\) is continuous at \(3\).
Practice 6
From the left,
\[ \lim_{x\to2^-}f(x)=5. \]From the right,
\[ \lim_{x\to2^+}f(x)=5. \]Therefore,
\[ \boxed{k=5}. \]Practice 7
At the left endpoint \(0\),
\[ \lim_{x\to0^+}\sqrt{x}=0=f(0). \]Therefore, \(\sqrt{x}\) is continuous from the right at \(0\).
Practice 8
\[ x^2-4=(x-2)(x+2). \]The function is undefined at \(x=-2\) and \(x=2\).
Therefore, it is continuous on
\[ \boxed{(-\infty,-2)\cup(-2,2)\cup(2,\infty)}. \]Practice 9
No. Continuity requires
\[ \lim_{x\to5}f(x)=f(5). \]Here,
\[ 7\neq10. \]Therefore, \(f\) is not continuous at \(5\).
Practice 10
- \(f(a)\) is defined.
- \(\lim_{x\to a}f(x)\) exists.
- \(\lim_{x\to a}f(x)=f(a)\).
What to Remember
Continuity at a point is summarized by
\[ \boxed{\lim_{x\to a}f(x)=f(a)}. \]- The function value must exist.
- The two-sided limit must exist.
- The limit must equal the function value.
- On closed intervals, endpoints use one-sided continuity.
- Polynomials are continuous everywhere.
- Rational functions are continuous wherever their denominators are nonzero.
- Piecewise formulas must be checked where their definitions change.
- A removable hole can be repaired by assigning the limiting value.
The central test is simple: does the function's actual value agree with what the nearby values are approaching?
Where We Go Next
When continuity fails, the failure can occur in several different ways. Some discontinuities are removable, some involve jumps, and others involve unbounded or oscillatory behavior.
The next lesson, Types of Discontinuity, classifies these failures and connects each type to its corresponding limit behavior.