AP Calculus AB and BC

Limit of (e^x - 1 - x)/x^2 at 0 Is 1/2

The limit of e to the x minus 1 minus x, over x squared, as x approaches 0 is one half. Subtracting the first two terms of the exponential series leaves x squared over 2 as the leading behaviour, so dividing by x squared gives one half.

lim⁡x→0ex−1−xx2=12\lim_{x \to 0} \frac{e^{x}-1-x}{x^{2}} = \frac{1}{2}

Settled by L'Hopital's rule twice, or the Maclaurin series.

Why subtracting two terms raises the order

ex=1+x+x22+x36+⋯e^{x} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \cdots

Removing 11 and xx deletes exactly the terms that would dominate, leaving x22\frac{x^{2}}{2} as the largest survivor. Dividing by x2x^{2} then gives 12\frac{1}{2}.

lim⁡x→0ex−1−xx2  =H  lim⁡x→0ex−12x  =H  lim⁡x→0ex2=12\lim_{x \to 0}\frac{e^{x}-1-x}{x^{2}} \;\overset{\text{H}}{=}\; \lim_{x \to 0}\frac{e^{x}-1}{2x} \;\overset{\text{H}}{=}\; \lim_{x \to 0}\frac{e^{x}}{2} = \frac{1}{2}

A family worth recognising

Each extra subtracted term raises the order by one, so the matching denominator power changes with it.

  • ex−1x→1\frac{e^{x}-1}{x} \to 1
  • ex−1−xx2→12\frac{e^{x}-1-x}{x^{2}} \to \frac{1}{2}
  • ex−1−x−x22x3→16\frac{e^{x}-1-x-\frac{x^{2}}{2}}{x^{3}} \to \frac{1}{6}

The pattern is 1n!\frac{1}{n!}, which is exactly the coefficient the series predicts.

The mistakes students make

  • Stopping after one pass and reporting 00 from ex−12x\frac{e^{x}-1}{2x}.
  • Assuming the subtraction makes the numerator vanish faster than the denominator. Both are order 22, which is why the answer is finite and nonzero.

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 - 1 - x)/x^2 as x approaches 0?

It is 12\frac{1}{2}.

Why subtract 1 and x?

Those are the first two terms of the exponential series, so removing them leaves x22\frac{x^{2}}{2} as the leading behaviour.