CEGEP Calculus II — Improper Integrals and Indeterminate Forms Worksheet

All 6 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

1. Q1Indeterminate Products and Differences

   Evaluate ${\displaystyle {lim}_{x\to 0^{+}}{x}^2\ln x}$.

   1. Name the indeterminate form, and explain in one sentence why L'Hospital's rule cannot be applied to the product as it stands.
   2. Rewrite the product as a quotient and evaluate the limit.
2. Q2Indeterminate Powers

   Evaluate each limit by taking logarithms. In each case name the indeterminate form first.

   1. ${\displaystyle {lim}_{x\to \infty }{(1+\frac{3}{x})}^{2x}}$
   2. ${\displaystyle {lim}_{x\to 0^{+}}{x}^{3x}}$
3. Q3Improper Integrals over an Infinite Interval

   Consider ${\displaystyle {\int }_1^{\infty }\frac{\ln x}{x^3}dx}$.

   1. Write the integral as a limit of a proper definite integral.
   2. Evaluate it, and say what the value means about the region lying between the curve $y={\displaystyle \frac{\ln x}{x^3}}$ and the $x$-axis for $x\ge 1$.
4. Q4Improper Integrals with an Infinite Discontinuity

   A cylindrical tank of depth $4$ m drains through a hole in its base. Under one model the time to empty, in seconds, is ${\displaystyle T_1={\int }_0^4\frac{2}{\sqrt{h}}dh}$, where $h$ is the depth of water still in the tank; a second model gives instead ${\displaystyle T_2={\int }_0^4\frac{2}{h}dh}$.

   1. Explain why both integrals are improper, and write each as an explicit limit.
   2. Evaluate each, and say what the two results predict about the tank.
5. Q5Deciding Whether an Improper Integral Converges

   For each integral below, state which feature makes it improper, then decide whether it converges or diverges and justify your decision. A comparison with a simpler integral is enough wherever it settles the matter; you are not asked to evaluate anything a comparison already decides.

   1. ${\displaystyle {\int }_1^{\infty }\frac{dx}{x^2+e^x}}$
   2. ${\displaystyle {\int }_1^{\infty }\frac{x+3}{x^2+1}dx}$
   3. ${\displaystyle {\int }_0^1\frac{dx}{x+\sqrt{x}}}$
6. Q6Synthesis — drawing on several topics in this set

   Consider ${\displaystyle {\int }_0^1\frac{\ln x}{\sqrt{x}}dx}$.

   1. Explain why the integral is improper, and write it as an explicit limit.
   2. Evaluate ${\displaystyle {lim}_{a\to 0^{+}}\sqrt{a}\ln a}$, naming the indeterminate form and showing the rewrite that makes L'Hospital's rule applicable.
   3. Use parts (a) and (b) to decide whether the integral converges, and give its value if it does.

The 5 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

Which indeterminate forms belong to this course

L'Hospital's rule is split across the two calculus courses by the shape of the form, and that split is why this set looks the way it does. Calculus I takes the two quotient forms, 0/0 and ±∞/±∞, where the rule applies to the expression exactly as it is written. Calculus II takes the ones where it does not: ∞ − ∞, 0 · ∞, 1^∞, 1^−∞, ∞⁰ and (0⁺)⁰. Not one of those is a quotient, and the rule is a theorem about quotients — so each has to be rewritten into a quotient first, by putting a difference over a common denominator, by sending one factor of a product into the denominator, or by taking logarithms. That rewrite is the whole lesson here; the differentiating that follows it is the part you already own.

Q1 makes that the explicit task: name the form of x²ln x as x → 0⁺, then say in one sentence why the rule cannot be applied to the product as it stands. The honest sentence is not "the rule fails here". It is that the rule says nothing at all about an expression of this shape — its hypotheses are about a numerator and a denominator, and there is no denominator yet. Part (b) supplies one.

Powers go through the logarithm

Write that chain down in full every time, including the closing line. The middle step is where a power form becomes a product form, which is why step 3 above costs two rewrites: from power to product, then from product to quotient. Q2 part (a) has a base tending to 1 with an exponent growing without bound, part (b) a base tending to 0⁺ with an exponent tending to 0, and the two travel the same route.

An improper integral is a limit — that is the definition, not a formality

The symbol ∫ from 1 to ∞ is not a Riemann integral with a large upper endpoint. Riemann integration is defined on a bounded interval for a bounded function, and neither condition holds here, so the symbol is defined to mean a limit of proper integrals. Q3, Q4 and Q6 each ask you to write that limit out before evaluating anything, and the reason is not ceremony: the limit is the object you are asked about, and "converges" means precisely that this limit exists and is a finite number.

The two rows are the two ways an integral goes improper, and Q5 opens by asking you to say which one you are looking at in each case — an unbounded interval, or an unbounded integrand. They are found by different inspections. The first you see in the limits of integration; the second only by checking where the integrand is undefined and what it does as you approach that point. An integral can be improper for both reasons at once, in which case it is split into pieces that are each improper in one way, and it converges only if every piece converges on its own.

Q4: same interval, same bad point, and the exponent decides

Q4 is a draining tank under two models whose integrals differ only in the power of h in the denominator. Both are improper at the lower endpoint — the integrand is undefined at h = 0 and unbounded as h → 0⁺ — and nothing in the notation announces that, which is the point of asking you to explain why each is improper before you touch it. Two integrals that look nearly identical do not have to behave alike, and part (b) asks what the results predict about the tank: a finite time is a tank that empties, an infinite one is a model in which the last of the water never leaves. That physical reading is where a divergent integral stops being a technicality.

Q5: deciding convergence without evaluating

Q5 gives three integrals and asks for a verdict with a justification — and tells you outright not to evaluate anything a comparison already settles. That instruction is doing real work: one of the three integrands has no elementary antiderivative at all, so evaluation is not merely slower, it is unavailable. Comparison is the tool that answers the question actually asked, which is whether a number exists, not what it is.

The comparison functions worth knowing by heart are the ones Q5 reaches for: an exponential e^(−x) over an unbounded interval, and a simple root or power of x near a singular endpoint. Notice also that "this behaves like 1/x for large x" is a conjecture, not a proof. It tells you which verdict to aim at; the mark is earned by the inequality that follows, and manufacturing that inequality — bounding a denominator above to bound the fraction below, for instance — is the actual skill Q5 is testing.

Q6: where the two halves of the sheet meet

Q6 is the reason these topics share a worksheet. It is an improper integral with an infinite discontinuity at an endpoint, and when you integrate by parts and push the endpoint to its limit, the boundary term that appears is a product of something tending to 0 with a logarithm tending to −∞ — an indeterminate form of exactly the kind part (b) has you evaluate on its own, by the rewrite from the top of this page. Part (c) then makes the convergence question depend on that value. Improper integrals are not a separate topic that happens to sit next to indeterminate forms; they generate them, at the endpoint, every time a logarithm or a negative power is involved.

Getting the most out of it

Write the limit down before you integrate

One line, at the top: the integral equals the limit as t → ∞, or as the endpoint approaches the bad point from the correct side. Do this before finding an antiderivative, not after. It forces you to identify what is improper and where, and it is the line a marker looks for first, because it is the only evidence that you know what the symbol means.

Name the form before you differentiate anything

Substitute the target value mentally and write down what you get — 0 · ∞, ∞ − ∞, 1^∞ and so on. The name tells you which rewrite you need, it stops you applying L'Hospital's rule to something that is not a quotient, and it stops you applying it to something that was never indeterminate in the first place. Then name the form again after each rewrite and each application.

Check the whole interval, not just its ends

Every zero of a denominator, every place a logarithm meets 0, every negative power: locate them, then ask whether each lies inside the interval, at an endpoint, or outside it altogether. Only the last case lets you use the Fundamental Theorem directly. This one check is the difference between a correct verdict and a confident wrong number.

Answer in words, not just in symbols

The expected answer to a convergence question is a sentence: the limit exists and is finite, so the integral converges, or the limit is infinite, so the integral diverges. Writing "= ∞" as though it were a value is the habit that eventually produces an arithmetic with infinities, which is exactly the mistake the ∞ − ∞ form exists to punish.

Common questions