Evaluating 0/0 Limits by Taylor Expansion to the Leading Power

Idea

A limit of the form $\dfrac{f(x)}{g(x)}$ at $x\to 0$ that gives the indeterminate $\tfrac{0}{0}$ is decided entirely by the lowest-order nonzero terms of the numerator and denominator. Instead of differentiating repeatedly (L'Hopital), expand both functions as Taylor (Maclaurin) series around $0$ and compare the leading powers.

If near $x=0$ $f(x) = a_p\,x^{p} + O(x^{p+1}), \qquad g(x) = b_q\,x^{q} + O(x^{q+1}),$ with $a_p, b_q \neq 0$, then $\lim_{x\to 0}\frac{f(x)}{g(x)} = \begin{cases} 0, & p > q,\\[2pt] \dfrac{a_p}{b_q}, & p = q,\\[2pt] \pm\infty, & p < q. \end{cases}$ The whole problem reduces to finding the first surviving power of each and the coefficient sitting on it.

Method

1. Recognize $\tfrac00$. Confirm numerator and denominator both vanish at the target point (shift to $0$ first if needed).
2. Expand to a guessed order. Substitute the standard series for each building block, keeping terms up to some order $N$.
3. Find the leading power. Cancel terms that vanish; the lowest power that does not cancel is the order $p$ (numerator) and $q$ (denominator).
4. Re-expand if you cancelled too early. If the leading terms cancelled and you only kept $O(x^{N})$ with $N$ too small, go back and keep one more order.
5. Read off the ratio $a_p/b_q$ (only when $p=q$).

Toolbox: standard Maclaurin series

For a variable exponent, rewrite a power as an exponential first: $u^{v} = e^{v\ln u}$, expand the exponent, then exponentiate using the $e^{x}$ series.

Mini-example

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

• Numerator: $e^{x} - 1 - x = \left(1 + x + \tfrac{x^2}{2} + \cdots\right) - 1 - x = \tfrac{x^2}{2} + O(x^3)$, so $p=2$, $a_2=\tfrac12$.
• Denominator: $1 - \cos x = 1 - \left(1 - \tfrac{x^2}{2} + \cdots\right) = \tfrac{x^2}{2} + O(x^4)$, so $q=2$, $b_2=\tfrac12$.
• Ratio: equal powers, so the limit is $\dfrac{a_2}{b_2} = \dfrac{1/2}{1/2} = 1.$

Why it beats L'Hopital

When the leading powers are high (here both are $x^2$, and in the power-tower question both are $t^2$), L'Hopital forces two or more rounds of messy product/chain differentiation. One Taylor expansion to the matching order delivers every coefficient at once.

Exam trick: Expand only as far as the first nonzero power, but be ready to go one order deeper if the leading terms cancel. The classic trap is stopping at $O(t^2)$, seeing $0/0$ again, and panicking — keep the next term and the ratio appears immediately.