AP Calculus AB and BC

Limit of (1 + 2/x)^x at Infinity Is e^2

Q: Why does the 2 become an exponent?

Taking logarithms gives x\ln\left(1+\frac{2}{x}\right) \to 2, and exponentiating turns that into e^{2}.

The limit of 1 plus 2 over x, all to the x, as x approaches infinity, is e squared. The general compound interest form 1 plus k over x to the x tends to e to the k, so the constant inside becomes the exponent on e.

\lim_{x \to \infty} \left(1+\frac{2}{x}\right)^{x} = e^{2}

Settled by the compound interest limit with k = 2.

Logarithms make it routine

Set $y = \left(1+\frac{2}{x}\right)^{x}$ and take logarithms to bring the exponent down.

\ln y = x\ln\left(1+\frac{2}{x}\right) \longrightarrow 2 \implies y \to e^{2}

The step is the same one every $1^{\infty}$ problem uses: take a logarithm, resolve the resulting product, then exponentiate at the end.

The compound interest reading

This is $200$ percent annual interest compounded $x$ times a year. As the compounding gets finer the balance approaches $e^{2} \approx 7.389$ times the principal rather than growing without bound.

The mistakes students make

Answering $e$ and ignoring the $2$ .
Answering $2e$ . The constant becomes an EXPONENT, not a multiplier.
Forgetting to exponentiate at the end and reporting $2$ .

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 (1 + 2/x)^x at infinity?

It is $e^{2} \approx 7.389$ .

Why does the 2 become an exponent?

Taking logarithms gives $x\ln\left(1+\frac{2}{x}\right) \to 2$ , and exponentiating turns that into $e^{2}$ .