AP Calculus AB and BC

Limit of (1+1/x)^x as x Approaches Infinity Is e

Q: What is \lim_{x \to \infty}\left(1+\frac{k}{x}\right)^x?

It is e^k. The logarithm becomes x\ln\left(1+\frac{k}{x}\right), which tends to k by the same difference quotient, so the expression tends to e^k. With k = 1 this returns e, and with k = -1 it gives e^{-1}.

Q: Is this the compound interest formula?

It is the heart of it. An investment at annual rate r compounded n times a year multiplies by \left(1+\frac{r}{n}\right)^n, and letting n \to \infty gives e^r. Continuous compounding is exactly this limit, which is why e shows up in growth and decay models.

Q: Is 1^{\infty} ever simply 1?

Not as a limit form. If a base is exactly 1 then every power of it is 1, but 1^{\infty} as a limit describes a base that is only approaching 1. Depending on the speeds involved the answer can be 1, e, any positive number, or \infty, so the form has to be resolved rather than read off.

The limit of (1 + 1/x)^x as x approaches infinity is e, about 2.71828. Direct substitution gives the indeterminate form 1 to the infinity, which is not 1: the base shrinks toward 1 while the exponent grows without bound, and the two effects balance. This limit is one standard definition of e.

\lim_{x \to \infty} \left(1+\frac{1}{x}\right)^x = e

Settled by the definition of e.

Taking a logarithm to settle the race

The exponent is where the trouble lives, so move it down with a natural logarithm and see what that quantity does.

\ln\left[\left(1+\frac{1}{x}\right)^x\right] = x\ln\left(1+\frac{1}{x}\right) = \frac{\ln\left(1+\frac{1}{x}\right)}{\frac{1}{x}}

Write $h = \frac{1}{x}$ , so $h \to 0$ as $x \to \infty$ . The quotient becomes $\frac{\ln(1+h)}{h}$ , which is the difference quotient for $\ln(1+t)$ at $t = 0$ , and its limit is the derivative there: $\frac{1}{1+0} = 1$ .

\lim_{x \to \infty} x\ln\left(1+\frac{1}{x}\right) = \lim_{h \to 0} \frac{\ln(1+h)}{h} = 1

The logarithm of the expression approaches $1$ , so the expression itself approaches $e^1 = e$ . Read the other way, this limit is where $e$ comes from: $2.718281828\ldots$ has no tidier closed form, and many courses take this limit as its definition, which is why the argument above can look like it is going in a circle.

Why substituting infinity says nothing

Letting $x$ grow sends $\frac{1}{x}$ to $0$ , so the base slides down to $1$ while the exponent climbs to $\infty$ .

\left(1+\frac{1}{x}\right)^x \longrightarrow 1^{\infty}

$1^{\infty}$ is an indeterminate form, not the number $1$ . Powers of exactly $1$ are always $1$ , but this base is never exactly $1$ : it is $1 + \frac{1}{x}$ , always a little larger, and a number slightly above $1$ raised to a huge power can land anywhere.

Two forms, two different answers

Both $\left(1+\frac{1}{x^2}\right)^x$ and $\left(1+\frac{1}{x}\right)^{x^2}$ read as $1^{\infty}$ , yet the first tends to $1$ and the second to $\infty$ . The form alone decides nothing. What decides the value is how fast the base approaches $1$ compared with how fast the exponent grows, and only the logarithm exposes that.

The mistakes students make

Answering $1$ because the base tends to $1$ . That reasoning would hold if the exponent were a fixed number, and it is not.
Answering $\infty$ because the exponent tends to $\infty$ . That would hold if the base stayed pinned above $1$ , and it does not.
Applying L'Hopital's rule to the expression as written. The rule accepts $\frac{0}{0}$ and $\frac{\infty}{\infty}$ only, so the logarithm has to convert the power into a quotient first.
Taking the logarithm, getting $1$ , and reporting $1$ . The limit of the logarithm is $1$ ; the limit asked for is $e^1$ .

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

What is $\lim_{x \to \infty}\left(1+\frac{k}{x}\right)^x$ ?

It is $e^k$ . The logarithm becomes $x\ln\left(1+\frac{k}{x}\right)$ , which tends to $k$ by the same difference quotient, so the expression tends to $e^k$ . With $k = 1$ this returns $e$ , and with $k = -1$ it gives $e^{-1}$ .

Is this the compound interest formula?

It is the heart of it. An investment at annual rate $r$ compounded $n$ times a year multiplies by $\left(1+\frac{r}{n}\right)^n$ , and letting $n \to \infty$ gives $e^r$ . Continuous compounding is exactly this limit, which is why $e$ shows up in growth and decay models.

Is $1^{\infty}$ ever simply $1$ ?

Not as a limit form. If a base is exactly $1$ then every power of it is $1$ , but $1^{\infty}$ as a limit describes a base that is only approaching $1$ . Depending on the speeds involved the answer can be $1$ , $e$ , any positive number, or $\infty$ , so the form has to be resolved rather than read off.