Limits and Derivatives: Application Problems (9)

Question

Evaluate $\displaystyle \lim_{x \to 0} \frac{e^x - 1 - x - \frac{x^2}{2}}{x^3}$.

Solution — Step by Step

Final answer: $\frac{1}{6}$.

Why This Works

L’Hôpital’s rule converts $0/0$ or $\infty/\infty$ into the limit of derivatives. Each application reduces the order of the singularity by one. Three applications were needed because the denominator was $x^3$.

The numerator $e^x - 1 - x - x^2/2$ is exactly the Taylor remainder at order 3 — it equals $x^3/6 + O(x^4)$. Dividing by $x^3$ gives $1/6$ in the limit, matching our L’Hôpital answer.

Alternative Method (Faster)

Use the Taylor series of $e^x$:

Numerator becomes $\frac{x^3}{6} + \frac{x^4}{24} + \ldots$. Divide by $x^3$:

Three lines, no L’Hôpital. JEE Advanced expects you to know Taylor expansions of $e^x$, $\sin x$, $\cos x$, $\ln(1+x)$ at least up to fourth order.