Putting Continuity and the IVT Together
A continuity problem is often a sequence of decisions. First identify where the function is defined and continuous. Then decide whether the conclusion you want follows from continuity alone or requires the Intermediate Value Theorem (IVT). In particular, a sign change guarantees a zero only when the function is continuous throughout the interval between the tested points.
This workshop brings those decisions together. It covers how to classify common discontinuities, how to check continuity on an interval, and how to write a complete existence argument. In each case, the key is to state the relevant conditions before using a theorem.
Continuity and Common Discontinuities
If a function is not continuous at \(c\), the limits from the left and right, the function value, or both may fail to agree. Three useful classifications are:
- Removable discontinuity: the finite two-sided limit exists, but either \(f(c)\) is undefined or it differs from that limit. Defining or redefining the value at \(c\) to equal the limit repairs the discontinuity.
- Jump discontinuity: the finite left-hand and right-hand limits exist but are unequal.
- Infinite discontinuity: function values become unbounded near \(c\), such as when a one-sided limit is \(+\infty\) or \(-\infty\).
These labels describe what happens at a point. To decide whether a function is continuous on an interval, the point-by-point condition must hold everywhere in that interval, with one-sided continuity used at included endpoints.
Worked Example: Classifying Three Discontinuities
Classify the behavior at the indicated points for the following functions.
First, let \(f(x)=\frac{x^2-1}{x-1}\) for \(x\ne 1\), with \(f(1)=3\). For \(x\ne1\), factor the numerator:
Therefore \(\lim_{x\to1}f(x)=1+1=2\), while \(f(1)=3\). The limit exists and is finite, but does not equal the function value. This is a removable discontinuity. If instead we defined \(f(1)=2\), the limit and function value would agree, repairing it.
Next, let \(g(x)=\frac{1}{x-2}\). As \(x\to2^-\), the denominator is negative and approaches zero, so \(\lim_{x\to2^-}g(x)=-\infty\). As \(x\to2^+\), the denominator is positive and approaches zero, so \(\lim_{x\to2^+}g(x)=+\infty\). The function has an infinite discontinuity at \(2\).
Finally, let \(h(x)=-1\) for \(x<0\), and \(h(x)=1\) for \(x\ge0\). Then \(\lim_{x\to0^-}h(x)=-1\) and \(\lim_{x\to0^+}h(x)=1\). The one-sided limits are finite but unequal, so \(h\) has a jump discontinuity at \(0\). In particular, the two-sided limit does not exist, regardless of the value \(h(0)=1\).
Check: The classification follows directly from the limits: the first finite limit disagrees with the value, the second is unbounded, and the third has unequal finite one-sided limits.
Testing Continuity on an Interval
For a formula defined by different expressions on different parts of its domain, check each formula where it applies and pay particular attention to the joining points. For a rational function, denominator zeros require investigation because the formula is not defined there. A simplification valid away from such a point does not by itself assign the missing function value.
At an included endpoint, only the limit from within the interval is relevant. For example, to check continuity at the left endpoint \(a\) of \([a,b]\), compare \(f(a)\) with \(\lim_{x\to a^+}f(x)\). At an interior joining point, both one-sided limits must agree with each other and with the function value.
Worked Example: Checking Continuity Across an Interval
Define \(q\) on \([-2,2]\) by \(q(x)=\frac{x^2-1}{x-1}\) for \(x\ne1\), and \(q(1)=2\). Determine whether \(q\) is continuous on the entire interval.
For every \(x\ne1\), factor and cancel:
This expression is continuous away from \(1\). At the exceptional point, compute the limit and compare it with the assigned value:
Thus \(q\) is continuous at \(1\), as well as at every other interior point. At the left endpoint, \(\lim_{x\to-2^+}q(x)=\lim_{x\to-2^+}(x+1)=-1=q(-2)\). At the right endpoint, \(\lim_{x\to2^-}q(x)=\lim_{x\to2^-}(x+1)=3=q(2)\).
All interior points and both endpoints satisfy the appropriate continuity condition. Therefore \(q\) is continuous on \([-2,2]\). As a comparison, if the assigned value were \(q(1)=4\), the limit would remain \(2\), so the function would not be continuous on the interval.
Check: On the interval, \(q(x)=x+1\) at every point, including \(1\), because the separately assigned value is also \(2\). The endpoint values and one-sided limits agree.
When the Intermediate Value Theorem Applies
The closed interval and continuity throughout it are essential hypotheses. When the goal is to prove that an equation has a solution, it is often useful to write the equation as \(f(x)=0\), then check continuity and evaluate at two endpoints. For an equation comparing two expressions, subtract one side from the other: equality is equivalent to the difference being zero.
Worked Example: Proving a Polynomial Equation Has a Solution
Show that \(x^3+x-1=0\) has a solution in \((0,1)\).
Set \(f(x)=x^3+x-1\). A polynomial is continuous for every real \(x\), so \(f\) is continuous on \([0,1]\). Evaluate the endpoints exactly:
and
Since \(-1<0<1\), the value \(0\) lies between \(f(0)\) and \(f(1)\). The IVT therefore gives some \(c\in[0,1]\) such that \(f(c)=0\). Because \(f(0)=-1\ne0\) and \(f(1)=1\ne0\), this \(c\) is not either endpoint; hence \(c\in(0,1)\). By the definition of \(f\), \(f(c)=0\) means \(c^3+c-1=0\).
Check: Substituting the defining expression back into the conclusion gives \(f(c)=c^3+c-1=0\), exactly the required equation. The argument proves existence; it does not claim that the solution is unique.
Why an Auxiliary Difference Works
Suppose the goal is to prove that two continuous functions \(u\) and \(v\) take the same value somewhere on \([a,b]\). Define \(H(x)=u(x)-v(x)\). Since differences of continuous functions are continuous, \(H\) is continuous on the interval. Further, \(H(c)=0\) is equivalent to \(u(c)-v(c)=0\), which is equivalent to \(u(c)=v(c)\). If the endpoint values of \(H\) have opposite signs, the IVT gives precisely such a point.
Worked Example: Proving Two Graphs Intersect
Show that the graphs of \(u(x)=x^2\) and \(v(x)=2-x\) intersect at some point in \([0,2]\).
Define their difference \(H(x)=u(x)-v(x)\). Both \(u\) and \(v\) are polynomials, so they are continuous, and their difference is continuous on \([0,2]\). Simplify and evaluate at the endpoints:
Since \(-2<0<4\), the IVT gives \(c\in[0,2]\) with \(H(c)=0\). The endpoint values of \(H\) are nonzero, so \(c\in(0,2)\). Now check what this equation means:
Thus the graphs intersect at the input \(c\). As an additional algebraic check, \(H(x)=x^2+x-2=(x+2)(x-1)\), so \(H(1)=0\), and indeed \(u(1)=1\) and \(v(1)=1\). The IVT argument establishes existence; the factorization identifies an intersection in this particular example.
A Common Pitfall: Sign Change Without Continuity
A sign change by itself does not establish a zero. The IVT applies only if the function is continuous on the entire closed interval. For example, for the function \(h\) defined earlier, \(h(-1)=-1\) and \(h(1)=1\). A mistaken argument would say that the endpoint signs differ, so \(h(c)=0\) for some \(c\in[-1,1]\). But \(h(x)\) is always either \(-1\) or \(1\); it is never zero. The argument fails because \(h\) has a jump discontinuity at \(0\), so it is not continuous on \([-1,1]\).
Another common error is to cancel a factor and then assume the original function is continuous at the cancelled point. Cancellation establishes equality only where the cancelled factor is nonzero. At the point where it is zero, check the original function value and the limit separately, as in the continuity example above.
A Reliable Decision Process
For mixed continuity and existence questions, the following sequence keeps the reasoning precise:
Decide whether you must classify a point, establish continuity on an interval, prove a zero exists, or prove two functions agree somewhere.
Inspect denominator zeros and piecewise joins. Compare one-sided limits and function values; at included endpoints, use the one-sided limit from within the interval.
For a root, use the left side minus the right side so the desired equation becomes a zero. For an intersection, subtract one continuous function from the other.
State that the chosen function is continuous on the whole closed interval, and calculate its endpoint values exactly.
Use the IVT to obtain an input where the auxiliary function has the target value, then explain why that equation gives the solution or intersection required.
Check Your Understanding
For each question, identify the relevant continuity condition or IVT hypothesis before drawing a conclusion.
- If \(\lim_{x\to c}f(x)=5\) and \(f(c)=5\), what can you conclude about continuity at \(c\)?
- If the finite left- and right-hand limits at a point exist but differ, how is the discontinuity classified?
- What limit condition must be checked at the left endpoint of a closed interval?
- A function is continuous on \([a,b]\) and has \(f(a)<0<f(b)\). What does the IVT guarantee, and where does the resulting point lie?
- Why does opposite endpoint sign alone fail to prove a zero for a function with a jump inside the interval?
