Q.Evaluate the following limit:

[!FORMULA] $\lim_{x\to2}\dfrac{x^4-16}{x-2}$

From Guessing to Computing

Tables and graphs are fine for building intuition, but they are impractical as a general method — you cannot tabulate your way through an exam. This concept collects the theorems that let you compute a limit algebraically, in a "mechanical" way, without ever touching a table.

Direct substitution for polynomials

Theorem 9.1. If $P(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n$ is a polynomial, then for any real $x_0$,

In other words: for a polynomial, the limit is just direct substitution. This works because a polynomial has no "trouble points" — it's defined and smooth everywhere, so there's nothing for a limit to disagree with. (A constant function is the degree-0 special case: $\lim_{x\to x_0} c = c$.)

The algebra of limits

Theorem 9.2. Suppose $\lim_{x\to x_0} f(x)$ and $\lim_{x\to x_0} g(x)$ both exist, and $c$ is a constant. Then all of the following exist, and:

• Constant multiple: $\displaystyle\lim_{x\to x_0} c\,f(x) = c \lim_{x\to x_0} f(x)$
• Sum/Difference: $\displaystyle\lim_{x\to x_0} [f(x) \pm g(x)] = \lim_{x\to x_0} f(x) \pm \lim_{x\to x_0} g(x)$
• Product: $\displaystyle\lim_{x\to x_0} [f(x)\,g(x)] = \lim_{x\to x_0} f(x) \cdot \lim_{x\to x_0} g(x)$
• Quotient: $\displaystyle\lim_{x\to x_0} \frac{f(x)}{g(x)} = \frac{\lim_{x\to x_0} f(x)}{\lim_{x\to x_0} g(x)}$, provided $\lim_{x\to x_0} g(x) \neq 0$.

These rules extend to any finite number of functions being added, multiplied, etc.

The power rule

Theorem 9.3. If $\lim_{x\to x_0} f(x)$ exists, then for any positive integer (in fact any rational) $n$,

This is really just the product rule applied $n$ times to $f\cdot f \cdots f$.

The difference-of-powers limit

Theorem 9.4. For a positive integer $n$ (and in fact for any rational exponent),

Why it's true, in brief: factor $x^n - a^n = (x-a)(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \cdots + a^{n-1})$. Cancel the common factor $(x-a)$ (legitimate since we only ever consider $x \neq a$ when taking the limit) to leave $n$ terms, each of which tends to $a^{n-1}$ as $x \to a$ — giving a total of $n\,a^{n-1}$. This single result is a shortcut generator: it turns an entire family of "$0/0$"-looking quotients into instant answers, and — though its proof waits for a later chapter — it is secretly the seed of the derivative of $x^n$.

The three evaluation techniques for a $0/0$ form at a finite point

When direct substitution into a quotient produces $\frac{0}{0}$, the quotient theorem is blocked (the denominator's limit is $0$), so you cannot stop — you must first rewrite the expression so the offending factor cancels.

1. Factor and cancel. If both numerator and denominator vanish at $x_0$, then $(x - x_0)$ is a common factor of both — pull it out and cancel.

Fresh illustration: $\displaystyle\lim_{x\to 2} \frac{x^2 - x - 2}{x - 2}$. Direct substitution gives $\frac{0}{0}$. Factor the numerator: $x^2-x-2 = (x-2)(x+1)$. Cancel $(x-2)$ to get $\lim_{x\to 2}(x+1) = 3$.

2. Rationalise using the conjugate surd. When a square root sits in the numerator or denominator, multiply top and bottom by the conjugate to turn the root into a difference of squares, which then cancels the offending factor.

Fresh illustration: $\displaystyle\lim_{x\to 0} \frac{\sqrt{x+4}-2}{x}$. Multiply by $\dfrac{\sqrt{x+4}+2}{\sqrt{x+4}+2}$: the numerator becomes $(x+4)-4 = x$, so the expression becomes $\dfrac{x}{x(\sqrt{x+4}+2)} = \dfrac{1}{\sqrt{x+4}+2}$, and the limit is $\dfrac{1}{4}$.

3. Direct substitution — used whenever the function is a polynomial or, more generally, whenever the denominator's limit is not zero at the point in question (a rational function is continuous away from the zeros of its denominator, so substitution is valid there — see Concept 5).

Putting several of these theorems together handles compound expressions too: e.g. $\lim_{x\to 2}(x^3 - 3x + 6)(-x^2+15)$ splits via the product rule into $\lim_{x\to 2}(x^3-3x+6) \cdot \lim_{x\to 2}(-x^2+15)$, each evaluable by Theorem 9.1, and the two results multiplied.