Once you understand the basic idea of a limit, the next step is learning how to compute limits efficiently. Limit laws are a set of rules that let you break complicated limits into simpler ones — just like arithmetic rules for numbers, but for limits.
📌 Assumption
For all the laws below, we assume that $\displaystyle \lim_{x \to a} f(x)$ and $\displaystyle \lim_{x \to a} g(x)$ both exist (are finite numbers).
📋 The Limit Laws (Reference Table)
Sum Law
$\displaystyle \lim_{x \to a} [f(x) + g(x)] = L + M$
Difference Law
$\displaystyle \lim_{x \to a} [f(x) - g(x)] = L - M$
Constant Multiple Law
$\displaystyle \lim_{x \to a} [c \cdot f(x)] = c \cdot L$
Product Law
$\displaystyle \lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M$
Quotient Law
$\displaystyle \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M},\; M \neq 0$
Power Law
$\displaystyle \lim_{x \to a} [f(x)]^n = L^n$ (for integer $n$)
Root Law
$\displaystyle \lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{L}$ (for even $n$, $L \ge 0$)
Direct Substitution
If $f$ is a polynomial or rational function and $a$ is in the domain, then $\displaystyle \lim_{x \to a} f(x) = f(a)$
💡 Why These Matter
Instead of computing limits from scratch every time, you can use these laws to build up the limit of a complex expression from the limits of its simpler parts.
🔧 How to Use Limit Laws (Step by Step)
When you see a complicated limit, follow this strategy:
- Try direct substitution first — if you get a finite number, you're done.
- If you get $\frac{0}{0}$, try factoring, simplifying, or rationalizing.
- Break the expression into pieces using the sum, product, and quotient laws.
- Apply the constant multiple law to pull constants outside the limit.
- Use the power/root laws to handle exponents and roots.
Find $\displaystyle \lim_{x \to 2} (3x^2 - 4x + 1)$.
- $\displaystyle \lim_{x \to 2} (3x^2 - 4x + 1) = \lim_{x \to 2} 3x^2 - \lim_{x \to 2} 4x + \lim_{x \to 2} 1$ (Sum/Difference Law)
- $= 3 \cdot \lim_{x \to 2} x^2 - 4 \cdot \lim_{x \to 2} x + 1$ (Constant Multiple Law)
- $= 3 \cdot (4) - 4 \cdot (2) + 1 = 12 - 8 + 1 = 5$
Find $\displaystyle \lim_{x \to 1} (x^2 \cdot \sqrt{x})$.
- $\displaystyle \lim_{x \to 1} (x^2 \cdot \sqrt{x}) = \left(\lim_{x \to 1} x^2\right) \cdot \left(\lim_{x \to 1} \sqrt{x}\right)$ (Product Law)
- $= (1^2) \cdot \sqrt{1} = 1 \cdot 1 = 1$
Find $\displaystyle \lim_{x \to 3} \frac{x^2 - 9}{x - 3}$.
- Direct substitution gives $\frac{0}{0}$ — need to simplify first
- $\frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3} = x+3$ for $x \neq 3$
- Now $\displaystyle \lim_{x \to 3} (x+3) = 3 + 3 = 6$
Find $\displaystyle \lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$.
- Direct substitution gives $\frac{0}{0}$
- Multiply numerator and denominator by $\sqrt{x} + 2$: $\frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{(x-4)(\sqrt{x} + 2)} = \frac{x - 4}{(x-4)(\sqrt{x} + 2)} = \frac{1}{\sqrt{x} + 2}$ for $x \neq 4$
- Now $\displaystyle \lim_{x \to 4} \frac{1}{\sqrt{x} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}$
⚠️ Important: When Limit Laws Don't Apply
If any individual limit involved does not exist, you cannot apply these laws directly. For example, if $\lim f(x)$ does not exist, you cannot use the sum law to break it apart.
🔍 Key Takeaways
- Limit laws let you break complicated limits into simpler pieces.
- Always try direct substitution first — it works for polynomials and rational functions at points in their domain.
- If you get $\frac{0}{0}$, factor, simplify, or rationalize before applying limit laws.
- The quotient law requires the denominator's limit to be non‑zero.
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