Algebra of Limits

Once individual limits are known, they combine by simple arithmetic rules. Suppose $\lim_{x\to a} f(x) = L$ and $\lim_{x\to a} g(x) = M$, both existing (finite). Then:

• Sum / difference: $\displaystyle\lim_{x\to a}\big[f(x) \pm g(x)\big] = L \pm M$
• Scalar multiple: $\displaystyle\lim_{x\to a}\big[k\,f(x)\big] = k\,L$ for any constant $k$
• Product: $\displaystyle\lim_{x\to a}\big[f(x)\,g(x)\big] = L\,M$
• Quotient: $\displaystyle\lim_{x\to a}\dfrac{f(x)}{g(x)} = \dfrac{L}{M}$, provided $M \neq 0$
• Power: $\displaystyle\lim_{x\to a}\big[f(x)\big]^n = L^n$

Two limits are used constantly as building blocks: $\lim_{x\to a} k = k$ (a constant function) and $\lim_{x\to a} x = a$ (the identity function).

Direct substitution — when it is legal. For any polynomial $p(x)$, and for a rational function $\dfrac{p(x)}{q(x)}$ whenever the denominator's limit is non-zero, the algebra rules combine to give

So the first thing to try in any limit is plug in $x=a$. If it gives a definite number, that number is the limit. Only when substitution yields an indeterminate form such as $\tfrac{0}{0}$ must a technique (factoring, rationalising, a standard limit) be used.

Illustration. $\displaystyle\lim_{x\to 2}\frac{3x^2 + 1}{x + 4}$: the denominator's limit is $2+4 = 6 \neq 0$, so substitute directly: $\dfrac{3(4)+1}{6} = \dfrac{13}{6}$.