AP Calculus AB and BC

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

The limit of 1 minus 1 over x, all to the x, as x approaches infinity, is 1 over e. It is the compound interest limit with a negative rate: the general form 1 plus k over x to the x tends to e to the k, and here k is negative 1.

lim⁡x→∞(1−1x)x=1e\lim_{x \to \infty} \left(1-\frac{1}{x}\right)^{x} = \frac{1}{e}

Settled by the compound interest limit with k = -1.

The general form

lim⁡x→∞(1+kx)x=ek\lim_{x \to \infty}\left(1+\frac{k}{x}\right)^{x} = e^{k}

With k=−1k = -1 that gives e−1=1e≈0.368e^{-1} = \frac{1}{e} \approx 0.368. Taking logarithms proves it: xln⁡(1−1x)→−1x\ln\left(1-\frac{1}{x}\right) \to -1.

Why the form is indeterminate

The base tends to 11 and the exponent to infinity, which is the form 1∞1^{\infty}. It is indeterminate because the base approaches 11 from below at a rate that competes with the exponent's growth. Any answer between 00 and ∞\infty is possible depending on that race.

The mistakes students make

  • Answering 11 because the base tends to 11. That is exactly the indeterminate form.
  • Answering ee and missing the sign. A negative kk gives the reciprocal.

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 - 1/x)^x at infinity?

It is 1e≈0.368\frac{1}{e} \approx 0.368.

What is the general rule?

(1+kx)x→ek\left(1+\frac{k}{x}\right)^{x} \to e^{k} for any constant kk.