Applications of Series — Question 3

Question 3 – Solution

Step 1: Expand far enough to survive the cancellation.

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

Step 2: Subtract the specified terms.

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

Step 3: Divide by $x^2$.

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

Conclusion.

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