Reading limits from tables (Limits from Both Sides) works, but it is slow and never fully certain. The limit laws let us compute limits exactly by breaking a function into pieces whose limits we already know.
Key idea
If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then the limit of $f + g$ is $L + M$, of $f - g$ is $L - M$, of $c f$ is $cL$, of $fg$ is $LM$, of $f^n$ is $L^n$, and of $\dfrac{f}{g}$ is $\dfrac{L}{M}$ provided $M \ne 0$.
Why it works
Here is the idea for sums. Suppose $f(x)$ is within $0.01$ of $L$ and $g(x)$ is within $0.01$ of $M$. Then $f(x) + g(x)$ is within $0.01 + 0.01 = 0.02$ of $L + M$. Since we can make both errors as small as we like, we can make the error of the sum as small as we like too. The product and quotient laws follow the same plan with a little more algebra. The quotient law needs $M \ne 0$ because near $a$ we divide by $g(x)$, which must stay away from 0.
Direct substitution
Two simple limits start everything: $\lim_{x \to a} c = c$ and $\lim_{x \to a} x = a$. Build any polynomial from these with sums and products, and the laws say that $\lim_{x \to a} p(x) = p(a)$. For a rational function $\frac{p(x)}{q(x)}$, the same is true whenever $q(a) \ne 0$. So for these functions you may simply substitute, once you have checked the denominator.
Worked example: Combining known limits
Suppose $\lim_{x \to 4} f(x) = 3$ and $\lim_{x \to 4} g(x) = -2$. Find $\displaystyle\lim_{x \to 4} \frac{f(x)^2 + g(x)}{x - g(x)}$.
- Top: power and sum laws give $3^2 + (-2) = 7$.
- Bottom: $\lim (x - g(x)) = 4 - (-2) = 6$, which is not 0, so the quotient law applies.
- The limit is $\frac{7}{6}$.
Answer: $\frac{7}{6}$
Worked example: Substitution after a check
Find $\displaystyle\lim_{x \to 2} \frac{x^2 + 3x}{x + 1}$.
- Check the bottom first: $2 + 1 = 3 \ne 0$.
- So substitute: $\frac{4 + 6}{3} = \frac{10}{3}$.
Answer: $\frac{10}{3}$
- Can I substitute? Check the bottom first. If its limit is not 0, substitute and you are done.
- Bottom tends to 0 but top does not: expect $\pm\infty$ (an infinite limit) or a limit that does not exist.
- Both tend to 0: the form $\frac{0}{0}$ needs algebra first: factor, rationalize, or use a special limit.
Common mistake
The quotient law is used when the denominator's limit is 0, as in $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$. Writing $\frac{\lim (x^2 - 9)}{\lim (x - 3)} = \frac{0}{0}$ is not an answer; the law simply does not apply. Check: before dividing limits, always compute the limit of the bottom. If it is 0, you need another tool (factoring, in the next lesson).
The limit laws also hold for one-sided limits and for limits as $x \to \pm\infty$, as long as the limits involved are finite numbers.