Indeterminate forms and the number e

What you will be able to do

• name the indeterminate forms and say why each one needs work;
• turn $\infty - \infty$ and $0 \cdot \infty$ into forms you can handle;
• find limits of the type $1^{\infty}$ with the number $e$.

Why a form can be indeterminate

In the first course you met $\dfrac{0}{0}$. It does not have one value. Look at four limits that all look like $\dfrac{0}{0}$ at $x = 0$.

• $\dfrac{x}{x} \to 1$.
• $\dfrac{3x}{x} \to 3$.
• $\dfrac{x^2}{x} \to 0$.
• $\dfrac{x}{x^2} = \dfrac{1}{x}$, which has no limit.

The same shape gives four different answers. We say $\dfrac{0}{0}$ is indeterminate: the shape alone does not decide the limit. The same holds for these:

By contrast, $\dfrac{1}{0^+}$ is not indeterminate: it grows without bound. Also $\infty + \infty = \infty$, and $\infty \cdot \infty = \infty$.

Turning one form into another

$\infty - \infty$ by a common denominator. Find $\displaystyle\lim_{x \to 1}\left(\dfrac{1}{x - 1} - \dfrac{2}{x^2 - 1}\right)$. Both parts grow without bound. Combine them: $\dfrac{(x + 1) - 2}{x^2 - 1} = \dfrac{x - 1}{(x - 1)(x + 1)} = \dfrac{1}{x + 1}$, which tends to $\dfrac{1}{2}$.

$\infty - \infty$ with roots. Rationalise, as in the first course: $\sqrt{x^2 + x} - x \to \dfrac{1}{2}$.

$0 \cdot \infty$ by writing a quotient. Find $\displaystyle\lim_{x \to \infty} x\sin\dfrac{1}{x}$. The first factor grows and the second vanishes. Put $t = \dfrac{1}{x}$, so that $t \to 0$ and the expression is $\dfrac{\sin t}{t} \to 1$.

The number e as a limit

Put $1$ rupee in a bank that pays $100\%$ interest in a year. If the interest is added once, you have $2$ rupees. If it is added every half year, you have $\left(1 + \dfrac{1}{2}\right)^2 = 2.25$. If it is added $n$ times, you have $\left(1 + \dfrac{1}{n}\right)^n$. Does adding the interest more often make you rich without limit? No.

The values settle on $e \approx 2.71828$. So

The two are the same limit, with $x$ replaced by $\dfrac{1}{x}$. The form is $1^{\infty}$: the base tends to $1$, but the exponent tends to $\infty$.

Relatives of the limit

A change of variable gives a useful family.

Indeed $\left(1 + \dfrac{k}{x}\right)^x = \left[\left(1 + \dfrac{k}{x}\right)^{x/k}\right]^k$, and the inside tends to $e$.

A rule for $1^{\infty}$. If $f(x) \to 1$ and $g(x) \to \infty$, then

provided that the limit on the right exists. The reason is that $f^g = \left[(1 + (f - 1))^{1/(f - 1)}\right]^{g(f - 1)}$, and the inside tends to $e$.

Example. $\displaystyle\lim_{x \to \infty}\left(\dfrac{x + 3}{x + 1}\right)^{x + 4}$. Here $f - 1 = \dfrac{2}{x + 1}$, so $g(f - 1) = \dfrac{2(x + 4)}{x + 1} \to 2$. The limit is $e^2$.

Practise with new numbers

Try it in Python

The program prints $\left(1 + \dfrac{1}{n}\right)^n$ for $n = 1, 10, 100, \dots$ and compares $\left(1 + \dfrac{3}{n}\right)^n$ with $e^3$.

Common mistakes

• Saying $1^{\infty} = 1$. A base slightly above $1$ raised to a huge power can be large. The limit is $e$ to some power.
• Forgetting the exponent of $e$. For $\left(1 + \dfrac{k}{x}\right)^x$ the answer is $e^k$, not $e$.
• Saying $\infty - \infty = 0$. The difference of two large numbers can be anything. Combine or rationalise first.
• Applying the $1^{\infty}$ rule when the base does not tend to $1$. Check the base first.

Remember these

• Indeterminate forms: $\dfrac{0}{0}$, $\dfrac{\infty}{\infty}$, $\infty - \infty$, $0 \cdot \infty$, $1^{\infty}$.
• $\displaystyle\lim_{x \to \infty}\left(1 + \dfrac{1}{x}\right)^x = e$ and $\displaystyle\lim_{x \to 0}(1 + x)^{1/x} = e$.
• $\left(1 + \dfrac{k}{x}\right)^x \to e^k$ and $(1 + kx)^{1/x} \to e^k$.
• If $f \to 1$ and $g \to \infty$, then $f^g \to e^{\lim g(f - 1)}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find $\displaystyle\lim_{x \to \infty}\left(1 + \dfrac{5}{x}\right)^x$.

2. Find $\displaystyle\lim_{x \to 0}(1 + 3x)^{2/x}$.

3. Find $\displaystyle\lim_{x \to \infty} x\sin\dfrac{4}{x}$.

4. Find $\displaystyle\lim_{x \to 2}\left(\dfrac{1}{x - 2} - \dfrac{4}{x^2 - 4}\right)$.

5. Why is $\displaystyle\lim_{x \to \infty}\left(1 + \dfrac{1}{x^2}\right)^x$ equal to $1$ and not $e$?

References

• OpenStax, Calculus Volume 1, Chapter 4, section 4.6 on limits at infinity and section 4.8 on the indeterminate forms.
• NCERT, Mathematics Textbook for Class XI, Chapter 12, for the standard limits of polynomials and trigonometric functions. The limit of $\left(1 + \dfrac{1}{x}\right)^x$ is the definition of $e$ in most textbooks.