Algebraic Techniques

Resolving $\frac{0}{0}$ Indeterminate Forms with Algebraic Techniques

When evaluating $\lim_{x\to a} f(x)$ by direct substitution yields $\frac{0}{0}$, this does not mean the limit fails to exist. Instead, it signals that the expression contains hidden structure—typically a common factor or radical—that must be simplified before substitution works. The function is undefined exactly at $x=a$, but it may behave predictably as $x$ approaches $a$. The solution is to rewrite $f(x)$ algebraically into an equivalent form (valid for $x \ne a$) whose limit you can evaluate.

Three Core Techniques

Factoring: When numerator and denominator are polynomials both vanishing at $x=a$, they share the common factor $(x-a)$. Factor both, cancel the common factor, then substitute.

Example: $\lim_{x\to 2} \dfrac{x^2 - 4}{x-2} = \lim_{x\to 2} \dfrac{(x-2)(x+2)}{x-2} = \lim_{x\to 2}(x+2) = 4$

Rationalization: When square roots appear, multiply numerator and denominator by the conjugate. Use the identity $(\sqrt{u}-\sqrt{v})(\sqrt{u}+\sqrt{v}) = u - v$ to eliminate radicals.

Example: $\lim_{x\to 0} \dfrac{\sqrt{x+1}-1}{x} \cdot \dfrac{\sqrt{x+1}+1}{\sqrt{x+1}+1} = \lim_{x\to 0} \dfrac{x}{x(\sqrt{x+1}+1)} = \lim_{x\to 0} \dfrac{1}{\sqrt{x+1}+1} = \dfrac{1}{2}$

Common Denominators: For complex fractions, combine inner fractions first using a common denominator. This often reveals factors to cancel with outer terms.

Example: $\lim_{x\to 3} \dfrac{\frac{1}{x}-\frac{1}{3}}{x-3} = \lim_{x\to 3} \dfrac{\frac{3-x}{3x}}{x-3} = \lim_{x\to 3} \dfrac{-1}{3x} = -\dfrac{1}{9}$

The Strategy

Every technique follows the same pattern: rewrite the expression using algebra, simplify by canceling factors, then substitute to evaluate the limit. The key insight is that canceling a factor is valid for $x \ne a$, and limits depend only on behavior near $a$, not at $a$ itself.