Evaluating Limits — Substitution and Factorisation

When we need $\lim_{x\to a} f(x)$, the first thing to try is always direct substitution: simply put $x=a$ into $f(x)$. If $f$ is a polynomial, or any 'nice' combination of standard functions defined at $a$, this immediately gives the answer, by the algebra of limits above.

Direct substitution fails when it produces an indeterminate form such as $\dfrac{0}{0}$ — this does not mean the limit doesn't exist, only that this particular method cannot see it. The standard remedy for a $\dfrac{0}{0}$ form arising from a ratio of polynomials is factorisation: factor both numerator and denominator, cancel the common factor that is causing both to vanish at $x=a$, and then substitute.

For example, $\lim_{x\to2} \dfrac{x^2-4}{x-2}$ gives $\dfrac00$ on direct substitution. But $\dfrac{x^2-4}{x-2} = \dfrac{(x-2)(x+2)}{x-2} = x+2$ for every $x \neq 2$ — and since a limit only cares about values near $2$, not at $2$, we may cancel and then substitute: the limit is $2+2=4$.

When the indeterminate form arises from a surd (square root) expression, the analogous technique is rationalisation — multiply numerator and denominator by the conjugate of the surd expression to clear the $0/0$ pattern before substituting. …