MAT148H1 Chap.5 Infinite Limits, Continuity and Value Theorems
Infinite Limits, Continuity and Value Theorems
What an infinite limit actually claims
Saying lim x→a f(x) = ∞ is not a statement that the function equals infinity anywhere. It is a promise about heights: for every bound M > 0 there is a radius δ > 0 such that every x with 0 < |x − a| < δ produces f(x) > M. The structure mirrors the ε-δ definition of a finite limit, with the band |f(x) − L| < ε swapped for the half line f(x) > M.
A proof therefore starts by fixing an arbitrary M and ends by exhibiting a δ that depends only on M. For 1/x as x → 0⁺, choosing δ = 1/(M + 1) works, because 0 < x < δ forces 1/x > M + 1 > M.
An infinite limit is still a limit that fails to exist as a real number.
The difference is precision: proving nonexistence means negating the finite definition, showing that for every candidate L some ε > 0 defeats every δ. That argument usually needs cases on the sign of L, whereas the infinite limit proof needs only one choice of δ per M.
One-sided versions follow the same pattern.
For lim x→a⁺ f(x) = ∞ the window is a < x < a + δ, and for a limit of −∞ the conclusion becomes f(x) < −M. A vertical asymptote at x = a is present whenever at least one of these one-sided statements holds, so 1/x has an asymptote at 0 even though its two sides head in opposite directions.
Continuity and the three kinds of break
Take a function defined on an interval around a.
It is continuous at a when f(a) = lim x→a f(x). Three things are packed into that equation: f(a) exists, the limit exists, and the two agree. In ε-δ form, each ε > 0 has a matching δ > 0 so that |f(x) − f(a)| < ε holds for all x within δ of a. When continuity fails, the one-sided limits classify the damage.
Equal finite one-sided limits give a removable discontinuity, unequal finite ones give a jump, and a missing or infinite one-sided limit gives an essential discontinuity. Every vertical asymptote sits at an essential discontinuity.
The Squeeze Theorem is the standard tool for limits that algebra cannot reach directly. If f(x) ≤ g(x) ≤ h(x) near c and both outer functions tend to L, then g tends to L as well.
Oscillating factors such as sin(1/x) multiplied by a power of x are the classic targets: since −x² ≤ x² sin(1/x) ≤ x² and both bounds tend to 0, the middle function tends to 0 as x → 0.
Intermediate and Extreme Value Theorems
The Intermediate Value Theorem (IVT) states that if f is continuous on [a, b] and v lies strictly between f(a) and f(b), then f(x) = v for some x in (a, b).
It is non-constructive: it proves a solution exists without locating it. Its main use in proofs is root existence. For a polynomial, compute the endpoint values, observe a sign change, state continuity, and conclude.
Repeating the argument on half intervals (bisection) narrows the location as far as you like.
The Extreme Value Theorem (EVT) states that a function continuous on a closed interval [a, b] attains both a maximum and a minimum there. Both hypotheses carry weight. Drop closedness and x² on (0, 3) approaches 0 and 9 without reaching either.
Drop continuity and a function with a vertical asymptote inside [a, b] can be unbounded.
A good habit for EVT questions is to test each hypothesis separately. Ask whether the interval is closed, whether it is bounded, and whether the function is continuous at every point including the endpoints. If all three hold, the theorem guarantees the extremes.
If any fails, the theorem is silent, and you must either find the extremes directly or build explicit values showing they are missing.
Where these ideas lead next
Continuity is the hypothesis behind almost every later theorem in the course: the Mean Value Theorem needs continuity on [a, b], optimization on closed intervals relies on the EVT, and differentiability itself implies continuity.
Fluency with writing these quantified definitions now pays off when proofs of derivative rules and the Mean Value Theorem arrive in Units 3 and 5.
What this chapter covers
- 01
Infinite limits with M and δ
- 02
Limit does not exist versus limit equals infinity
- 03
Continuity at a point in ε-δ form
- 04
Removable, jump and essential discontinuities
- 05
Squeeze Theorem for oscillating limits
- 06
Intermediate Value Theorem root proofs
- 07
Extreme Value Theorem and its hypotheses
Worked example · free
Proving a root exists with the Intermediate Value Theorem
- 1State the hypothesis: q is a polynomial, hence continuous on ℝ and in particular on [1, 2].
- 2Compute endpoints: q(1) = 1 + 2 − 5 = −2 and q(2) = 8 + 4 − 5 = 7.
- 1Since −2 < 0 < 7, the IVT gives some c in (1, 2) with q(c) = 0.
- 2Bisect: q(1.5) = 3.375 + 3 − 5 = 1.375 > 0, while q(1) < 0, so a second application of the IVT on [1, 1.5] places a root in (1, 1.5).
Key terms
- Infinite Limit
- An infinite limit is the statement that f(x) exceeds every bound M once x is within some δ of a, with x ≠ a. It describes unbounded growth and is one specific way a finite limit can fail.
- Continuity at a Point
- Continuity at c is the condition that f is defined near c and its value f(c) matches the limit taken as x tends to c. Equivalently, small changes in x near c produce small changes in f(x).
- Essential Discontinuity
- An essential discontinuity is a break where at least one one-sided limit fails to exist or is infinite. Vertical asymptotes and wild oscillation such as sin(1/x) near 0 both produce this type.
- Non-constructive Theorem
- A non-constructive theorem is a result that guarantees an object exists without giving a method to find it. The Intermediate and Extreme Value Theorems are both of this kind, asserting existence only.
Infinite Limits, Continuity and Value Theorems FAQ
Is a limit that equals infinity the same as a limit that does not exist?
Every infinite limit is a nonexistent real limit, but the converse fails. Writing equals infinity records extra information: the values grow past every bound. A function like sin(1/x) near 0 has no limit and no infinite limit either. In a proof the two claims need different arguments.
How do I choose δ when proving a limit is infinity?
Work backwards from the target inequality f(x) greater than M. Solve for how close x must be to a, then set δ to that distance. For 1/(x − 3)² the condition (x − 3)² less than 1/M suggests δ = 1/√M. Check the choice by substituting back into the chain of inequalities.
Why does the Extreme Value Theorem need a closed interval?
On an open interval the function can approach an endpoint value without ever reaching it. The square function on (0, 3) gets arbitrarily close to 0 and 9 yet attains neither, so no maximum or minimum exists there. Closing the interval to [0, 3] restores both extremes.
Can the Intermediate Value Theorem find the exact root?
It only certifies that a root exists between two points with opposite signs. To approximate the root, repeat the sign test on smaller subintervals, which is bisection, or use later tools such as Newton's method. Each halving step needs only one new evaluation.
Exam move
Write the three quantified definitions (infinite limit, continuity, nonexistence) from memory on a blank page each day until they come out without hesitation. Then take two fresh polynomials, compute endpoint values, and write a complete IVT paragraph for each, naming continuity before any numbers.
Finish by inventing one counterexample for each EVT hypothesis and checking, with explicit values, that the maximum or minimum really is missing.
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