Indeterminate Forms and How to Resolve Them

When direct substitution produces an expression with no determinable value on its own, the result is called an indeterminate form. The forms met at this level are

"Indeterminate" does not mean "the limit does not exist" — it means substitution alone cannot decide the answer, so the expression must first be rewritten.

Distinguish an indeterminate form from a determinate one. $\tfrac{0}{0}$ is indeterminate (the answer could be anything). But $\tfrac{5}{0}$ (non-zero over zero) is not an indeterminate form — it signals the function is diverging, and the limit is $\pm\infty$ or does not exist; and $\tfrac{0}{5}$ is simply $0$. Only forms where two competing effects cancel are indeterminate.

The resolution toolkit — pick the tool that removes the offending factor:

1. Factor and cancel — for $\tfrac00$ from a common factor $(x-a)$, e.g. $\dfrac{x^2-9}{x-3} = x+3 \to 6$ as $x\to 3$.
2. Rationalise — for $\tfrac00$ involving a square-root difference, multiply by the conjugate.
3. Apply a standard limit — for $\tfrac00$ of trigonometric, exponential, or $\dfrac{x^n-a^n}{x-a}$ shape (§4).
4. Divide by the highest power — for $\tfrac{\infty}{\infty}$ from a rational function as $x\to\infty$ (§5).
5. Combine into a single fraction — for $\infty-\infty$, subtract over a common denominator to expose a $\tfrac00$ or $\tfrac{\infty}{\infty}$ form that a tool above then resolves.

Illustration — $\infty-\infty$. $\displaystyle\lim_{x\to 1}\left(\frac{1}{x-1} - \frac{2}{x^2-1}\right)$. Combining over the common denominator $x^2-1=(x-1)(x+1)$: $\dfrac{(x+1)-2}{(x-1)(x+1)} = \dfrac{x-1}{(x-1)(x+1)} = \dfrac{1}{x+1} \to \dfrac{1}{2}$. …