AP Calculus AB and BC

Limit of (1+x)^(1/x) as x Approaches 0 Is e

Q: Is this the same limit as \left(1+\frac{1}{x}\right)^x?

Yes, with the variable inverted. Setting u = \frac{1}{x} sends x \to 0^+ to u \to \infty and turns one expression into the other, so both equal e. Textbooks quote whichever version fits the problem in front of them.

Q: What is \lim_{x \to 0}(1+kx)^{1/x}?

It is e^k. Taking logs gives \frac{\ln(1+kx)}{x}, which tends to k, so the expression tends to e^k. For instance (1+2x)^{1/x} \to e^2 and (1-x)^{1/x} \to e^{-1}.

Q: Can I use L'Hopital's rule directly on (1+x)^{1/x}?

Not on the power itself, since the rule only handles \frac{0}{0} and \frac{\infty}{\infty} quotients. Take the logarithm first to reach \frac{\ln(1+x)}{x}, which is \frac{0}{0}; L'Hopital then gives \frac{1}{1+x} \to 1, and exponentiating returns e.

The limit of (1 + x)^(1/x) as x approaches 0 from the right is e, about 2.71828. Direct substitution gives 1 to the infinity, an indeterminate form. Substituting u = 1/x turns it into (1 + 1/u)^u with u going to infinity, which is the limit that defines e.

\lim_{x \to 0^+} (1+x)^{1/x} = e

Settled by the definition of e.

Turning it into the defining limit

Let $u = \frac{1}{x}$ . As $x \to 0^+$ the new variable $u$ runs to $\infty$ , and the expression rewrites itself into the standard form for $e$ .

(1+x)^{1/x} = \left(1+\frac{1}{u}\right)^{u}, \qquad u = \frac{1}{x} \to \infty

Working it directly instead, take a logarithm to bring the exponent down. The result is the difference quotient for $\ln(1+t)$ at $t = 0$ , whose value is the derivative $\frac{1}{1+0} = 1$ .

\ln\left((1+x)^{1/x}\right) = \frac{\ln(1+x)}{x} \longrightarrow 1

So the logarithm of the expression approaches $1$ , which puts the expression itself at $e^1 = e$ . Both routes are the same fact: $\left(1+\frac{1}{u}\right)^u$ and $(1+x)^{1/x}$ are one limit written with the variable inverted.

Why substituting zero says nothing

At $x = 0$ the base $1+x$ is $1$ and the exponent $\frac{1}{x}$ has no value at all. Approaching instead, the base slides toward $1$ from above while the exponent climbs without bound.

(1+x)^{1/x} \longrightarrow 1^{\infty}

$1^{\infty}$ is indeterminate. The base is never exactly $1$ , only near it, so the answer depends on a race between a base drifting toward $1$ and an exponent running to infinity. The logarithm is what turns that race into a $\frac{0}{0}$ quotient the standard tools can handle.

Does the left side agree?

It does, which is worth knowing because one-sided statements often warn that the other side misbehaves. For $-1 < x < 0$ the base $1+x$ is still positive, so the power is defined, and the same logarithm argument runs unchanged: $\frac{\ln(1+x)}{x}$ is a negative over a negative, so it stays positive, and it settles to $1$ from above.

The two sides close in on $1$ from opposite directions. On the right the quotient sits below $1$ and rises, about $0.9531$ at $x = 0.1$ ; on the left it sits above $1$ and falls, about $1.0536$ at $x = -0.1$ . That is why $(1+x)^{1/x}$ lands just under $e$ for $x > 0$ , about $2.5937$ at $x = 0.1$ , and just over $e$ for $x < 0$ , about $2.8680$ at $x = -0.1$ .

\lim_{x \to 0^-} (1+x)^{1/x} = \lim_{x \to 0^+} (1+x)^{1/x} = e

So the two-sided limit exists and is also $e$ . The right-hand version is the one usually written because it lines up with $u = \frac{1}{x} \to +\infty$ . The left side is perfectly legitimate, but it only reaches back to $x = -1$ , where the base $1+x$ stops being positive, so the substitution route runs out sooner there.

The mistakes students make

Answering $1$ from the base, or $\infty$ from the exponent. Neither factor gets to decide alone.
Treating $\frac{1}{x}$ as though it shrinks to $0$ along with $x$ . It is the reciprocal, so it grows without bound.
Reading $(1+x)^{1/x}$ as $1 + x^{1/x}$ . The exponent applies to the whole base, and $\lim_{x \to 0^+} x^{1/x}$ is $0$ , a different problem with a different answer.
Finishing at the logarithm. Getting $\frac{\ln(1+x)}{x} \to 1$ is the middle step; the answer 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

Is this the same limit as $\left(1+\frac{1}{x}\right)^x$ ?

Yes, with the variable inverted. Setting $u = \frac{1}{x}$ sends $x \to 0^+$ to $u \to \infty$ and turns one expression into the other, so both equal $e$ . Textbooks quote whichever version fits the problem in front of them.

What is $\lim_{x \to 0}(1+kx)^{1/x}$ ?

It is $e^k$ . Taking logs gives $\frac{\ln(1+kx)}{x}$ , which tends to $k$ , so the expression tends to $e^k$ . For instance $(1+2x)^{1/x} \to e^2$ and $(1-x)^{1/x} \to e^{-1}$ .

Can I use L'Hopital's rule directly on $(1+x)^{1/x}$ ?

Not on the power itself, since the rule only handles $\frac{0}{0}$ and $\frac{\infty}{\infty}$ quotients. Take the logarithm first to reach $\frac{\ln(1+x)}{x}$ , which is $\frac{0}{0}$ ; L'Hopital then gives $\frac{1}{1+x} \to 1$ , and exponentiating returns $e$ .