Theorems on limits

Building a table of values or sketching a graph every time a limit needs to be found is neither practical nor rigorous. This section collects a set of theorems — stated here without proof, since a full justification needs the formal $\varepsilon$-$\delta$ definition of a limit, beyond this book's scope (Theorem 9.4 is the one exception, whose proof is elementary algebra and is worked out below) — that let a limit be evaluated mechanically once certain simpler limits are known to exist.

Theorem 9.1 (limit of a polynomial by direct substitution). If $P(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n$ is a polynomial, then for any real $x_0$,

In other words, for a polynomial the limit is always found simply by plugging in $x_0$ — no algebraic trick is ever needed. (Example 9.7: $\lim_{x\to3}(x^3-2x+6)=27$ is just $P(3)$. Example 9.8 is the special case of a constant polynomial $P(x)=5$: the limit of a constant function is that same constant, for any $x_0$.)

Theorem 9.2 (algebra of limits). Suppose $\lim_{x\to x_0} f(x)$ and $\lim_{x\to x_0} g(x)$ both exist, and $c$ is a constant. Then each of $cf$, $f+g$, $f-g$, $fg$, and (provided $\lim_{x\to x_0} g(x)\ne0$) $f/g$ also has a limit at $x_0$, and

These extend to any finite number of functions. (Example 9.9 applies rule (i): $\lim_{x\to8}5x=5\cdot8=40$. Example 9.10 splits a sum into two separately-computable pieces and adds the results. Example 9.11 uses the power rule below to raise a linear-limit result to the tenth power rather than expanding a huge binomial by hand.)

Theorem 9.3 (power rule). If $\lim_{x\to x_0} f(x)$ exists, then so does $\lim_{x\to x_0}[f(x)]^n$, and

(Example 9.12 evaluates two separate polynomial limits and multiplies the results together, using this rule along the way to avoid expanding a cube directly.)

Theorem 9.4. $\displaystyle \lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$, and this stays true even when $n$ is any rational number, not merely a positive integer.

Proof. Factor the numerator using the standard difference-of-powers identity,

which has exactly $n$ terms inside the second bracket. Dividing both sides by $(x-a)$ cancels that factor — valid since the limit only ever considers $x\ne a$ — leaving

Now let $x\to a$: each of the $n$ terms on the right tends to $a^{n-1}$ (by Theorem 9.1, since each term is itself a polynomial in $x$), so the whole sum tends to $n$ copies of $a^{n-1}$ added together, i.e. $na^{n-1}$. $\blacksquare$ …