L'Hôpital's rule and improper integrals

The first week of Calculus II gives you two tools together. L’Hôpital’s rule solves indeterminate forms with derivatives. Improper integrals define, with limits, integrals whose limit is infinite or whose integrand blows up. The two are linked: when finding the value of an improper integral, an L’Hôpital limit almost always shows up at the end. The integral test in week 3 is also built directly on this topic.

The exam typically has 2-3 limits and 1-2 integrals of the “evaluate or show that it diverges” kind. Golden rules:

1. Plug in first. L’Hôpital is valid only for the forms $0/0$ or $\infty/\infty$.
2. Differentiate the numerator and denominator separately. Do not use the quotient rule.
3. Turn the other indeterminate forms ($0\cdot\infty$, $\infty-\infty$, $1^\infty$, $0^0$, $\infty^0$) into $0/0$ or $\infty/\infty$ first.
4. Improper integral = limit. You do not plug in infinity or the singular point directly; you write $b$ and take a limit.
5. If there is a singularity inside the interval, split the integral at that point. If one of the pieces diverges, the whole thing diverges.

L’Hôpital’s rule

The rule. Let $f$ and $g$ be differentiable near $a$, $g'(x)\neq 0$, and let $\lim_{x\to a} f(x)/g(x)$ be of the form $0/0$ or $\infty/\infty$. Then

$\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}$

(if the limit on the right exists or is $\pm\infty$). Here $a$ can be a finite number, $a^\pm$ or $\pm\infty$.

Intuition. In the $0/0$ case both functions go to zero. Their slopes decide which one goes faster. Near $a$, $f(x)\approx f'(a)(x-a)$ and $g(x)\approx g'(a)(x-a)$, and the ratio becomes $f'(a)/g'(a)$.

Example 1. $\displaystyle\lim_{x\to 0}\frac{e^x-1-x}{x^2}$

plug in x=0: (1-1-0)/0 = 0/0    → L'Hôpital
derivative:  (e^x - 1)/(2x)     → 0/0 again
derivative:  e^x / 2            → x=0: 1/2

The answer is $1/2$. Check the form again at every step; you stop when the form is no longer indeterminate.

PITFALL · check the form. $\displaystyle\lim_{x\to 0}\frac{x+1}{\cos x}$ is not indeterminate, it is directly $1/1 = 1$. If you apply L’Hôpital to it, you get $\dfrac{1}{-\sin x}$ and head to a wrong result.

Other indeterminate forms

Example 2. $\displaystyle\lim_{x\to 0^+} x\ln x$. The form is $0\cdot(-\infty)$.

$x\ln x = \frac{\ln x}{1/x} \;\Rightarrow\; \frac{1/x}{-1/x^2} = -x \to 0.$

Example 3. $\displaystyle\lim_{x\to\infty}\left(1+\frac{3}{x}\right)^x$. The form is $1^\infty$. Let $y$ be this expression:

$\ln y = x\ln\!\left(1+\frac3x\right) = \frac{\ln(1+3/x)}{1/x} \;\xrightarrow{\text{L'H}}\; \frac{\dfrac{-3/x^2}{1+3/x}}{-1/x^2} = \frac{3}{1+3/x}\to 3.$

So $\ln y\to 3$, $y\to e^3$. The general pattern: $\lim_{x\to\infty}(1+a/x)^{bx} = e^{ab}$.

Example 4. $\displaystyle\lim_{x\to 0^+} x^x$. The form is $0^0$. $\ln y = x\ln x \to 0$ (Example 2), so $y\to e^0 = 1$.

TIP · do not forget the last step. In $1^\infty$, $0^0$, $\infty^0$ questions, the limit you find is the limit of $\ln y$. The answer is $e^{(\text{that number})}$. This last line is where most points are lost.

Growth rates

As $x\to\infty$: $\ln x \ll x^p \ll e^x$ (for every $p>0$). L’Hôpital shows this: $\dfrac{\ln x}{x^p}\to\dfrac{1/x}{p x^{p-1}} = \dfrac{1}{p x^p}\to 0$. You will use this ordering constantly in the series chapter.

The exponential function beats every polynomial. For $\displaystyle\lim_{x\to\infty}\frac{x^2}{e^x}$, apply L’Hôpital twice: $\dfrac{2x}{e^x}$, then $\dfrac{2}{e^x}\to 0$. With each derivative the degree of the polynomial drops by one, while $e^x$ never changes. For $x^{10}$ you take ten derivatives, and the result is still $0$. In the exam, do not wonder “how many times should I apply L’Hôpital”; keep going while the form stays indeterminate, and stop when the indeterminacy is gone.

Sometimes L’Hôpital does not work and goes in a loop: when $\dfrac{\sqrt{x^2+1}}{x}$ is differentiated it becomes $\dfrac{x}{\sqrt{x^2+1}}$, and if you differentiate again you are back at the start. In that case use algebra: factor out the largest power in the numerator and denominator, $\sqrt{1+1/x^2}\to 1$.

Improper integrals

The definite integral $\int_a^b f\,dx$ is defined for a bounded function on a finite interval. This breaks down in two cases:

Type 1: infinite interval.

$\int_a^\infty f(x)\,dx = \lim_{b\to\infty}\int_a^b f(x)\,dx, \qquad \int_{-\infty}^\infty f\,dx = \int_{-\infty}^c f\,dx + \int_c^\infty f\,dx.$

Type 2: the integrand goes to infinity at the boundary or inside. If $f$ blows up at the point $b$, then

$\int_a^b f(x)\,dx = \lim_{c\to b^-}\int_a^c f(x)\,dx.$

If the limit is a finite number the integral converges, otherwise it diverges. For $\int_{-\infty}^{\infty}$ and for cases with a singularity inside, each piece must converge separately.

Example 5. $\displaystyle\int_1^\infty \frac{dx}{x^2} = \lim_{b\to\infty}\left[-\frac1x\right]_1^b = \lim_{b\to\infty}\left(1-\frac1b\right) = 1.$

Example 6. $\displaystyle\int_1^\infty \frac{dx}{x} = \lim_{b\to\infty}\ln b = \infty$, diverges. $1/x$ goes to zero, but not fast enough. The area under it is infinite.

Example 7. $\displaystyle\int_0^1 \frac{dx}{\sqrt x} = \lim_{a\to 0^+}\left[2\sqrt x\right]_a^1 = 2.$ The function goes to infinity but the area is finite.

Example 8. $\displaystyle\int_0^\infty x e^{-x}\,dx$. Integration by parts: $\int x e^{-x}dx = -xe^{-x} - e^{-x}$.

$\lim_{b\to\infty}\left(-be^{-b}-e^{-b}\right) - (0 - 1) = 0 + 1 = 1.$

The step $b e^{-b} = b/e^b \to 0$ is an L’Hôpital limit.

p-integrals

The reference families you will use for comparison:

$p=1$ diverges on both sides. Memorize this table.

Comparison tests

Even if you cannot compute the integral, you can show whether it converges.

Direct comparison. Let $0\le f(x)\le g(x)$ for $x\ge a$.

• If $\int_a^\infty g$ converges, then $\int_a^\infty f$ converges too.
• If $\int_a^\infty f$ diverges, then $\int_a^\infty g$ diverges too.

Limit comparison. If $f,g>0$ and $\lim_{x\to\infty} f(x)/g(x) = L$ with $0<L<\infty$, the two integrals either both converge or both diverge.

Example 9. $\int_1^\infty e^{-x^2}dx$ cannot be computed (with elementary functions), but for $x\ge1$, $x^2\ge x$, so $e^{-x^2}\le e^{-x}$. Since $\int_1^\infty e^{-x}dx = 1/e$ converges, $\int_1^\infty e^{-x^2}dx$ converges.

Example 10. $\displaystyle\int_1^\infty \frac{dx}{\sqrt{x}+1}$. Limit comparison with $g = 1/\sqrt x$: $\dfrac{1/(\sqrt x+1)}{1/\sqrt x} = \dfrac{\sqrt x}{\sqrt x+1}\to 1$. $\int_1^\infty x^{-1/2}dx$ diverges ($p=1/2\le1$), so the given integral diverges too.

Singularity inside

Example 11. $\displaystyle\int_0^3 \frac{dx}{(x-1)^{2/3}}$. The integrand blows up at $x=1$, and this point is inside the interval $[0,3]$. Split in two:

$\int_0^1 (x-1)^{-2/3}dx = \lim_{c\to1^-}\left[3(x-1)^{1/3}\right]_0^c = 0-3(-1) = 3,$

$\int_1^3 (x-1)^{-2/3}dx = \lim_{c\to1^+}\left[3(x-1)^{1/3}\right]_c^3 = 3\sqrt[3]{2}-0.$

Both pieces are finite, the total is $3+3\sqrt[3]{2}\approx 6.78$. If one of the pieces had come out infinite, the answer would be “diverges”; the value of the other would not matter.

Solution template for the exam

In an “evaluate or show that it diverges” question, follow this order:

1) Find the problem points: is a limit ∞? is there an x that makes the denominator zero? is ln or a root outside its domain?
2) Write a separate piece for each problem point, and write each piece with lim
3) Find the antiderivative (u-sub, integration by parts, partial fractions)
4) Take the limit; use L'Hôpital if needed (like b e^-b, ln b / b)
5) All finite → converges, write the value.  One is ∞ → diverges

If the antiderivative cannot be found (for example $e^{-x^2}$, $\dfrac{1}{\sqrt{x^4+1}}$), the question already asks “converge or diverge?”. Then instead of steps 3 and 4, you do a comparison: look at the behavior of the integrand as $x\to\infty$, find the dominant term, and compare with a p-integral. For example $\dfrac{1}{\sqrt{x^4+1}}\approx \dfrac{1}{x^2}$, so it converges.

PITFALL · the direction of the comparison. “I am bigger than something that converges” proves nothing. To show convergence you must show that you are smaller than something that converges, and to show divergence that you are bigger than something that diverges.

Common mistakes