AP Calculus AB and BC

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

The limit of e to the x minus e, over x minus 1, as x approaches 1 is e. The expression is the alternate form of the derivative of e to the x at x equals 1, and since that function is its own derivative the value is e.

lim⁡x→1ex−ex−1=e\lim_{x \to 1} \frac{e^{x}-e}{x-1} = e

Settled by recognising it as a derivative at a point.

A derivative in disguise

Since e1=ee^{1} = e, the numerator is f(x)−f(1)f(x) - f(1) with f(x)=exf(x) = e^{x}, so the quotient is exactly f′(1)f'(1).

lim⁡x→1ex−ex−1=ddxex∣x=1=e1=e\lim_{x \to 1}\frac{e^{x}-e}{x-1} = \left.\frac{d}{dx}e^{x}\right|_{x=1} = e^{1} = e

Spotting the shape saves the problem

Any limit of the form f(x) minus f(a) over x minus a is a derivative at a. Recognising it converts a limit you cannot see into a differentiation you can do in one line.

Factoring also works

Writing ex−e=e(ex−1−1)e^{x} - e = e\left(e^{x-1}-1\right) and substituting u=x−1u = x-1 turns it into e⋅eu−1ue \cdot \frac{e^{u}-1}{u}, and that standard limit is 11, leaving ee.

The mistakes students make

  • Cancelling the ee terms. The ee in the numerator is not a factor of the whole expression as written.
  • Reporting that the limit does not exist because substitution gives 00\frac{0}{0}.

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 (e^x - e)/(x - 1) at 1?

It is ee.

Why is it e?

The expression is the derivative of exe^{x} at x=1x = 1, and exe^{x} is its own derivative.