AP Calculus AB and BC

Limit of x ln(1 + 1/x) at Infinity Is 1

Q: How does it relate to e?

Exponentiating gives \left(1+\frac{1}{x}\right)^{x} \to e. This limit is that one seen through a logarithm.

The limit of x times the natural log of 1 plus 1 over x, as x approaches infinity, is 1. Substituting u equals 1 over x turns it into the standard limit of ln of 1 plus u over u. It is the logarithm of the limit that defines e.

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

Settled by substitution to a standard logarithm limit.

Substitute to a known limit

Let $u = \frac{1}{x}$ , so $u \to 0^{+}$ as $x \to \infty$ , and the product becomes a quotient.

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

This is the log of the e limit

Exponentiating both sides recovers the compound interest limit: (1 + 1/x)^x tends to e. The two facts are the same statement, one seen through a logarithm.

Why the form is indeterminate

The first factor runs to infinity while the second runs to $0$ , so the product is the indeterminate form $\infty \cdot 0$ . Converting it into a quotient is the standard move, and it is what makes L'Hopital applicable if you prefer that route.

The mistakes students make

Answering $0$ because the logarithm tends to $0$ , or $\infty$ because $x$ does. The product is indeterminate.
Splitting $\ln\left(1+\frac{1}{x}\right)$ into $\ln 1 + \ln\frac{1}{x}$ . There is no rule for a logarithm of a sum.

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 the limit of x ln(1 + 1/x) at infinity?

It is $1$ .

How does it relate to e?

Exponentiating gives $\left(1+\frac{1}{x}\right)^{x} \to e$ . This limit is that one seen through a logarithm.