Algebra of limits and evaluating limits

Once you know two limits, you can compute many more by combining them. The algebra of limits tells you exactly how. Combined with a few standard tricks (factor, cancel, rationalise), it lets you handle every limit problem in this chapter.

The limit laws

Assume $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$. Then:

1. Sum/difference. $\lim_{x \to a} [f(x) \pm g(x)] = L \pm M$.
2. Scalar. $\lim_{x \to a} [c \cdot f(x)] = c L$.
3. Product. $\lim_{x \to a} [f(x) g(x)] = L M$.
4. Quotient. $\lim_{x \to a} [f(x)/g(x)] = L/M$, provided $M \ne 0$.
5. Power. $\lim_{x \to a} [f(x)]^n = L^n$.
6. Root. $\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{L}$, provided $L \ge 0$ for even $n$.

Direct substitution

For continuous functions, direct substitution always works: $\lim_{x \to a} f(x) = f(a) \quad \text{if } f \text{ is continuous at } a.$

All polynomials, rational functions (where the denominator is nonzero), exponentials, sine, cosine, square roots (where defined) are continuous. Try direct substitution first. If you get a number, you're done. If you get $0/0$, you need a trick.

The $0/0$ form

When substitution gives $0/0$, the limit is indeterminate , it could be anything (or fail to exist). Try:

Factor. $\lim_{x \to 1} \dfrac{x^2 - 1}{x - 1} = \lim_{x \to 1} (x + 1) = 2$.

Rationalise. $\lim_{x \to 0} \dfrac{\sqrt{1 + x} - 1}{x}$. Multiply by $\dfrac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1}$: $\lim_{x \to 0} \frac{(1 + x) - 1}{x(\sqrt{1 + x} + 1)} = \lim_{x \to 0} \frac{1}{\sqrt{1 + x} + 1} = \frac{1}{2}.$

Use a standard limit. Identify $\sin x / x$, $(a^x - 1)/x$, etc. (Next subtopic.)

The $\infty/\infty$ form (as $x \to \infty$)

For rational functions as $x \to \infty$, divide by the highest power of $x$ in the denominator: $\lim_{x \to \infty} \frac{3 x^2 + x - 1}{2 x^2 - 5} = \lim_{x \to \infty} \frac{3 + 1/x - 1/x^2}{2 - 5/x^2} = \frac{3}{2}.$

In general:

• If degree numerator < degree denominator: limit $= 0$.
• If equal: limit $=$ ratio of leading coefficients.
• If degree numerator > degree denominator: limit $= \pm\infty$.

Worked examples

Example 1. $\lim_{x \to 2} (x^3 - 4 x + 1)$.

Direct substitution: $8 - 8 + 1 = 1$.

Example 2. $\lim_{x \to 3} \dfrac{x^2 - 9}{x^2 - 4 x + 3}$.

Substitute: $0/0$. Factor: $\dfrac{(x-3)(x+3)}{(x-1)(x-3)} = \dfrac{x+3}{x-1}$. Substitute now: $\dfrac{6}{2} = 3$.

Example 3. $\lim_{x \to 0} \dfrac{\sqrt{4 + x} - 2}{x}$.

Rationalise: $\dfrac{(\sqrt{4+x} - 2)(\sqrt{4+x} + 2)}{x(\sqrt{4+x} + 2)} = \dfrac{x}{x(\sqrt{4+x}+2)} = \dfrac{1}{\sqrt{4+x} + 2}$. As $x \to 0$: $\dfrac{1}{4}$.

Example 4. $\lim_{x \to a} \dfrac{x^n - a^n}{x - a}$.

This is the slope of $f(x) = x^n$ at $x = a$, and equals $n a^{n-1}$ (a standard limit). Factor: $x^n - a^n = (x - a)(x^{n-1} + x^{n-2} a + \dots + a^{n-1})$ (geometric-style factoring), so the ratio equals $x^{n-1} + x^{n-2} a + \dots + a^{n-1}$. At $x = a$: $n \cdot a^{n-1}$.

Example 5. $\lim_{x \to \infty} \dfrac{5 x^3 - 2 x + 7}{2 x^3 + 4}$.

Divide by $x^3$: $\dfrac{5 - 2/x^2 + 7/x^3}{2 + 4/x^3} \to \dfrac{5}{2}$.

Try it yourself

1. $\lim_{x \to 1} (3 x^2 + 2 x - 1)$.
2. $\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2}$.
3. $\lim_{x \to -1} \dfrac{x^2 + 3 x + 2}{x + 1}$.
4. $\lim_{x \to 0} \dfrac{\sqrt{1 + 2 x} - 1}{x}$.
5. $\lim_{x \to 3} \dfrac{x^3 - 27}{x - 3}$.
6. $\lim_{h \to 0} \dfrac{(2 + h)^2 - 4}{h}$.
7. $\lim_{x \to \infty} \dfrac{2 x^2 + 3}{x^2 - 1}$.
8. $\lim_{x \to \infty} \dfrac{4 x + 7}{x^2 + 1}$.
9. $\lim_{x \to a} \dfrac{\sqrt{x} - \sqrt{a}}{x - a}$ (for $a > 0$).
10. $\lim_{x \to 1} \dfrac{x^4 - 1}{x^2 - 1}$.
11. $\lim_{x \to 0} \dfrac{(x + 5)^2 - 25}{x}$.
12. $\lim_{x \to 4} \dfrac{x - 4}{\sqrt{x} - 2}$.

Pitfalls / Tricks

• Always try direct substitution first.
• For $0/0$, factor, cancel, or rationalise.
• $\infty/\infty$: divide by the highest power.
• Insight. A "$0/0$" is the indeterminate form that calculus is built to resolve , limits exist exactly because we can compute these meaningfully.