Applications of Series — Question 3

*Question 3 Evaluate $\displaystyle\lim_{x\to0}\frac{e^x-1-x}{x^2}$ by using a Taylor expansion. Identify why the constant and linear terms disappear and which coefficient determines the limit.

*Question 3 – Solution

$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4).$$

$$e^x-1-x=\frac{x^2}{2}+\frac{x^3}{6}+O(x^4).$$ The constant $1$ and linear term $x$ cancel exactly.

$$\frac{e^x-1-x}{x^2}=\frac12+\frac{x}{6}+O(x^2).$$ As $x\to0$, every remaining positive-power term tends to zero.

$$\boxed{\displaystyle\lim_{x\to0}\frac{e^x-1-x}{x^2}=\frac12}.$$ The answer is the first Taylor coefficient not removed by the numerator’s subtraction.