Limit of $\frac{e^{x} - 1 - x}{x^{2}}$ as $x \to 0$

How to work it out

1. 1

   We want what (exp(x) - 1 - x) / $x^{2}$ approaches as x → 0.

   $$\lim_{x \to 0} \frac{e^{x} - 1 - x}{x^{2}}$$
2. 2

   Direct substitution gives an indeterminate form: 0/0. It does not tell us the limit.

3. 3

   L'Hôpital's rule: when the form is 0/0, the limit equals the limit of the ratio of the derivatives.

   $$\lim \frac{e^{x} - 1 - x}{x^{2}} = \lim \frac{e^{x} - 1}{2 x}$$
4. 4

   Still 0/0 (both vanish again), so apply it once more.

   $$\lim \frac{e^{x} - 1}{2 x} = \lim \frac{e^{x}}{2}$$
5. 5

   Now substitute x = 0: the denominator is not 0 any more.

   $$\frac{1}{2} = 0.5$$
6. 6

   Method: the series eˣ = 1 + x + x²/2 + ….

7. 7

   Result:

   $$\lim_{x \to 0} \frac{e^{x} - 1 - x}{x^{2}} = \frac12$$

Values of (exp(x) - 1 - x) / $x^{2}$ as x → 0 (computed)

Questions

More well-known limits

• $\frac{\sin\left(x\right)}{x}$ as $x \to 0$
• $\frac{\tan\left(x\right)}{x}$ as $x \to 0$
• $\frac{1 - \cos\left(x\right)}{x^{2}}$ as $x \to 0$
• $\frac{1 - \cos\left(x\right)}{x}$ as $x \to 0$
• $\frac{e^{x} - 1}{x}$ as $x \to 0$
• $\frac{\ln\left(1 + x\right)}{x}$ as $x \to 0$
• $\frac{\arctan\left(x\right)}{x}$ as $x \to 0$
• $\frac{\arcsin\left(x\right)}{x}$ as $x \to 0$