Evaluating Limits Algebraically

Direct Substitution

Try $x = a$ first. If it gives a real number, that's the limit.

Factoring

For $\frac{0}{0}$: factor and cancel. $\lim_{x \to 3} \frac{x^2-9}{x-3} = \lim_{x \to 3} \frac{(x+3)(x-3)}{x-3} = 6$.

Rationalizing (Conjugate)

For radical expressions: multiply by conjugate. $\lim_{x \to 0} \frac{\sqrt{x+4}-2}{x} \times \frac{\sqrt{x+4}+2}{\sqrt{x+4}+2} = \lim_{x \to 0} \frac{x}{x(\sqrt{x+4}+2)} = \frac{1}{4}$.

Common Factor

$\lim_{x \to 0} \frac{\frac{1}{x+2}-\frac{1}{2}}{x}$: combine fractions in numerator first.