Computing Limits Algebraically

Why this matters¶

A metal cube is heated and expands. On Polynomial and Rational Functions we asked how fast its volume grows when its side is exactly 2 centimetres. As the side grows from 2 to $x$ , the volume $x^3$ grows by $x^3 - 8$ , so on average it grows by

A(x) = \frac{x^3 - 8}{x - 2}

(1)

cubic centimetres per centimetre of side. At $x = 2$ the formula gives $\frac{0}{0}$ , which is not a number.

Figure 1:Graph of the average rate $y = \frac{x^3 - 8}{x - 2}$ for $0 \le x \le 4$ , rising from 4 at $x = 0$ to 28 at $x = 4$ , with an open circle at $x = 2$ , where the formula is undefined. The widget draws the circle at the height it estimates from the values on both sides of 2. A slider for $x$ moves a point along the graph. A table lists the values at $x = 1.9, 1.99, 1.999$ : 11.41, 11.9401, 11.994001, and at $x = 2.001, 2.01, 2.1$ : 12.006001, 12.0601, 12.61. They get closer to 12 from both sides.

Try this: move the point towards $x = 2$ from both sides and read the values. Then compute $x^2 + 2x + 4$ at $x = 1.99$ and at $x = 2$ . What do you notice?

On that page we factored the numerator, $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$ , cancelled the non-zero number $x - 2$ for $x \ne 2$ , and said that for $x$ close to 2 the rate is close to $2^2 + 2 \cdot 2 + 4 = 12$ . The table agrees. But two questions remain.

Is 12 the limit of $A(x)$ as $x \to 2$ ? The quotient law, part (d) of the limit laws, cannot say: the denominator approaches 0.
Cancelling changed the function. $A$ is undefined at 2, while $x^2 + 2x + 4$ is defined there. Why should the new function have the same limit as the old one?

This page answers both with one lemma: two functions that agree near $a$ , except possibly at $a$ itself, have the same limit at $a$ . With it, every limit of the form “ $\frac{0}{0}$ ” on this page follows the same plan: rewrite the function for $x \ne a$ , by factoring, rationalising or simplifying, until the limit laws apply, and let the lemma carry the limit back.

Limits of the form $\frac{0}{0}$ ¶

On Limit Laws we met quotients whose numerator and denominator both approach 0. The quotient law needs the limit of the denominator to be non-zero, so it does not apply to them.

Let $f$ and $g$ be functions, each defined at every point of an open interval containing $a$ , except possibly at $a$ , with $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = 0$ , and suppose that $\frac{f}{g}$ is also defined at every point of an open interval containing $a$ , except possibly at $a$ . We then say that $\lim_{x \to a} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ .

In words. “Of the form $\frac{0}{0}$ ” describes the situation, not the answer: substituting the limits of the parts gives the symbol $\frac{0}{0}$ , which is not a number, because 0 has no reciprocal. Such a limit may be any number, or not exist at all, as the figure on Limit Laws suggests and Example 3 proves. The form says only that the quotient law gives nothing, and that we must rewrite the function first.

Example. $\lim_{x \to 2} \frac{x^3 - 8}{x - 2}$ is of the form $\frac{0}{0}$ : by part (a) of direct substitution, the numerator approaches $2^3 - 8 = 0$ and the denominator $2 - 2 = 0$ , and the quotient is defined at every $x \ne 2$ .

Non-example. $\lim_{x \to 2} \frac{x^3 - 8}{x + 2}$ is not of the form $\frac{0}{0}$ : the denominator approaches $4 \ne 0$ , so the quotient law applies, and the limit is $\frac{0}{4} = 0$ . Nor is $\lim_{x \to 0} \frac{1}{x}$ , where the numerator approaches 1: that limit does not exist, as Limit Laws shows, and no rewriting changes that.

Main results¶

The lemma says that a limit only sees the values of a function on a small window around $a$ , with $a$ itself left out. The window is the set of $x$ with $0 < \abs{x - a} < r$ , which by part (e) of the proposition on distance inequalities is the open interval $(a - r, a + r)$ with its centre $a$ removed.

Proof

A window that wins a round of the $\eps$ – $\delta$ game for $g$ , made small enough to lie in the window where $f$ and $g$ agree, wins the same round for $f$ , because on it $f$ has the same values as $g$ .

By hypothesis, $f$ is defined at every $x$ with $0 < \abs{x - a} < r$ , which is the open interval $(a - r, a + r)$ with $a$ removed, by part (e) of the proposition on distance inequalities. So the definition of a limit applies to $f$ .

Let $\eps > 0$ . Since $\lim_{x \to a} g(x) = L$ , the definition gives a $\delta_0 > 0$ such that $g$ is defined and $\abs{g(x) - L} < \eps$ at every $x$ with $0 < \abs{x - a} < \delta_0$ . Let $\delta = \min(\delta_0, r)$ . It is one of the two positive numbers $\delta_0$ and $r$ , so it is positive.

Let $0 < \abs{x - a} < \delta$ . Since $\delta \le r$ and $\delta \le \delta_0$ , transitivity (property 2 of the order rules, in the form “ $u < v \le w$ implies $u < w$ ”) gives $\abs{x - a} < r$ and $\abs{x - a} < \delta_0$ . The first, with $0 < \abs{x - a}$ , says that $x$ is in the window of the hypothesis: $f$ is defined at $x$ , and $f(x) = g(x)$ . The second says that $\abs{g(x) - L} < \eps$ . So

\abs{f(x) - L} = \abs{g(x) - L} < \eps .

(3)

So $f$ is defined at every $x$ with $0 < \abs{x - a} < \delta$ , and $\delta$ wins the round $\eps$ for $f$ . Since $\eps > 0$ was arbitrary, $\lim_{x \to a} f(x) = L$ .

In words. If two functions have the same values near $a$ , except possibly at $a$ , they have the same limit at $a$ . What they do at $a$ plays no part: $f(a)$ and $g(a)$ may be different, or undefined. What they do far from $a$ plays no part either: only some window around $a$ , however small, matters.

Both directions. The hypothesis is the same with $f$ and $g$ swapped. So if either function has a limit at $a$ , the other has the same limit; and if one has no limit at $a$ , neither has the other.
A whole window is needed. Agreeing on one side of $a$ is not enough. The function $\frac{\abs{x}}{x}$ agrees with the constant 1 at every $x > 0$ , and the constant has the limit 1, but $\frac{\abs{x}}{x}$ has no limit at 0 (Exercise 7): every window around 0 also contains points $x < 0$ , where its value is ${-1}$ .
What we already knew. The remark on the value at $a$ says that changing a function at the single point $a$ , or leaving it undefined there, does not change its limit, and part (c) of direct substitution says that a rational function whose denominator is not 0 at $a$ , used only on an open interval around $a$ , keeps its limit $\frac{p(a)}{q(a)}$ . The lemma is the general form of both ideas. Unlike the remark, it asks $f$ and $g$ to agree only on a window around $a$ , not at every $x \ne a$ . Unlike part (c), it allows $g$ to be any function that has a limit at $a$ , given by any formula, and $f$ need not be defined at $a$ . This is what limits of the form $\frac{0}{0}$ need.

Example. For $x \ne 1$ , $\frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1} = x + 1$ , cancelling the non-zero number $x - 1$ . Both functions are defined at every $x$ with $0 < \abs{x - 1} < 1$ and agree there, and $\lim_{x \to 1} (x + 1) = 2$ by part (a) of direct substitution. By the lemma, $\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2$ .

Non-example. Agreeing at $a$ is not the hypothesis, and it is not enough. Let $h(x) = x + 1$ for $x \ne 0$ and $h(0) = 0$ . The functions $x$ and $h$ agree at $x = 0$ . But $\lim_{x \to 0} x = 0$ (part (a) of the limit laws), while $h$ agrees with $x + 1$ at every $x \ne 0$ , so by the lemma and part (a) of direct substitution, $\lim_{x \to 0} h(x) = 1$ .

The method for $\frac{0}{0}$ ¶

To find $\lim_{x \to a} f(x)$ for a quotient $f$ :

Substitute. Find the limits of the numerator and the denominator with the limit laws or direct substitution. If the denominator’s limit is not 0, the quotient law gives the answer, and we are done. If both limits are 0, the limit is of the form $\frac{0}{0}$ : go on.
Rewrite. Find a function $g$ and a window $0 < \abs{x - a} < r$ on which $f$ and $g$ are both defined and $f(x) = g(x)$ . The algebra may use $x \ne a$ : on the window we may cancel a factor that is not 0 there, such as $x - a$ .
Compute $\lim_{x \to a} g(x)$ with the limit laws or direct substitution, checking the hypotheses of each law and naming its part.
Conclude with Lemma 1: $\lim_{x \to a} f(x)$ is the same number.

Step 2 is where the three techniques below come in. Each of them produces a $g$ to which the laws apply.

Three ways to rewrite¶

Factoring. Let $f = \frac{p}{q}$ , with polynomials $p$ and $q$ such that $p(a) = 0$ and $q(a) = 0$ , and $p$ not the zero polynomial. (If $p$ is the zero polynomial, $f$ is 0 wherever it is defined; the rigorous track treats this case.) A polynomial of degree 0 is a non-zero constant, which has no root, so $p$ has degree at least 1. So has $q$ , which is not the zero polynomial, by the definition of a rational function. By part (b) of the factor theorem, for every real $x$ ,

\begin{aligned} p(x) &= (x - a)\, p_1(x), \\ q(x) &= (x - a)\, q_1(x), \end{aligned}

(4)

where $p_1$ and $q_1$ are polynomials whose degrees are one less than those of $p$ and $q$ (part (a) of the theorem, for degree at least 1). To find $p_1$ , divide $p$ by $x - a$ : since $p(x) = (x - a)\, p_1(x) + 0$ , the uniqueness in division of polynomials says that $p_1$ is the quotient, with the zero polynomial as the remainder. For a quadratic, a factorisation found by inspection is just as good, once multiplying out confirms it.

At every $x \ne a$ the number $x - a$ is not 0, so by part (a) of the sign rules, $q(x) \ne 0$ exactly when $q_1(x) \ne 0$ . So $\frac{p}{q}$ and $\frac{p_1}{q_1}$ are defined at the same points $x \ne a$ , and at those points

\frac{p(x)}{q(x)} = \frac{(x - a)\, p_1(x)}{(x - a)\, q_1(x)} = \frac{p_1(x)}{q_1(x)},

(5)

cancelling the non-zero factor $x - a$ . If $q_1(a) \ne 0$ , then by part (b) of the proposition on the domain of a rational function there is an open interval $(\alpha, \beta)$ containing $a$ at every point of which $\frac{p_1}{q_1}$ is defined. Let $r$ be the smaller of $a - \alpha$ and $\beta - a$ , which is positive, since $\alpha < a < \beta$ . If $0 < \abs{x - a} < r$ , then $a - r < x < a + r$ by part (a) of the proposition on distance inequalities, and $\alpha \le a - r$ and $a + r \le \beta$ (add $\alpha - r$ , or $a$ , to both sides of $r \le a - \alpha$ , or of $r \le \beta - a$ ; property 3 of the order rules), so $\alpha < x < \beta$ (property 2). So both functions are defined and agree on the window $0 < \abs{x - a} < r$ , part (b) of direct substitution gives $\lim_{x \to a} \frac{p_1(x)}{q_1(x)} = \frac{p_1(a)}{q_1(a)}$ , and the lemma gives the same limit for $\frac{p}{q}$ . If $q_1(a) = 0$ too, there are two cases.

If $p_1(a) = 0$ as well, the limit is again of the form $\frac{0}{0}$ , and we factor again (Exercise 5).
If $p_1(a) \ne 0$ , the limit does not exist, as for $\frac{x}{x^2}$ at 0 in Example 3 (iii), where $p_1(x) = 1$ and $q_1(x) = x$ .

The rigorous track, “Factoring always ends”, proves both cases: after finitely many steps, factoring gives the limit or shows that there is none. In an example, the window is usually found directly, from the roots of the denominator.

The cube of Why this matters is a special case of a general fact. For every polynomial $p$ and every real $a$ , part (a) of the factor theorem gives a polynomial $q$ such that, for every real $x \ne a$ ,

\frac{p(x) - p(a)}{x - a} = q(x) .

(6)

Both sides are defined at every $x \ne a$ and agree there, so on every window around $a$ . By part (a) of direct substitution, $\lim_{x \to a} q(x) = q(a)$ , and by the lemma

\lim_{x \to a} \frac{p(x) - p(a)}{x - a} = q(a) .

(7)

Rationalising. A difference of square roots becomes a difference of numbers when it is multiplied by the sum. For real numbers $u, v \ge 0$ , the numbers $\sqrt{u}$ and $\sqrt{v}$ are defined, and $\bigl(\sqrt{u}\bigr)^2 = u$ and $\bigl(\sqrt{v}\bigr)^2 = v$ , by the square-root remark. Multiplying out, the two middle terms cancel:

\bigl(\sqrt{u} - \sqrt{v}\bigr)\bigl(\sqrt{u} + \sqrt{v}\bigr) = u - v .

(8)

The conjugate $\sqrt{u} + \sqrt{v}$ is positive if $u > 0$ : then $\sqrt{u} \ge 0$ , and $\sqrt{u} \ne 0$ because $0^2 = 0 \ne u$ , so $\sqrt{u} > 0$ ; adding this to $\sqrt{v} \ge 0$ gives $\sqrt{u} + \sqrt{v} > 0$ (property 4 of the order rules). So we may multiply the numerator and the denominator of a quotient by it without changing the value: the quotient is the same number, now written with $u - v$ in place of $\sqrt{u} - \sqrt{v}$ (Example 2).

After rationalising, the limit of a square root comes from part (f) of the limit laws, the root law, and its hypotheses must be checked. If $\lim_{x \to a} h(x) = L$ with $L > 0$ , then $\lim_{x \to a} \sqrt{h(x)} = \sqrt{L}$ . If $L = 0$ , the law needs more: an $r > 0$ such that $h(x) \ge 0$ at every $x$ with $0 < \abs{x - a} < r$ . For example, $\abs{x} = \sqrt{x^2}$ for every real $x$ , by property 5 of the absolute value; here $h(x) = x^2$ has the limit 0 at 0 (part (a) of direct substitution), and $x^2 \ge 0$ for every real $x$ , by part (c) of the sign rules. So the root law gives

\lim_{x \to 0} \abs{x} = \lim_{x \to 0} \sqrt{x^2} = 0 .

(9)

Without the sign condition the law says nothing: $\sqrt{x}$ is undefined at every $x < 0$ , so it has no limit at 0 in the sense of the definition, although $x \to 0$ .

Simplifying compound fractions. A quotient whose numerator or denominator is itself a sum of fractions is brought to a single fraction first. For numbers $b \ne 0$ and $d \ne 0$ ,

\frac{c_1}{b} - \frac{c_2}{d} = \frac{c_1 d - c_2 b}{b d},

(10)

and $bd \ne 0$ by part (a) of the sign rules. Then a factor $x - a$ usually appears in the new numerator and cancels (Example 4).

Worked examples¶

Factoring¶

Example 1 (The cube from Why this matters)

Find $\displaystyle \lim_{x \to 2} \frac{x^3 - 8}{x - 2}$ , the rate at which the cube’s volume grows when its side is 2 cm.

Substitute. The function $A(x) = \frac{x^3 - 8}{x - 2}$ is defined at every $x \ne 2$ , since $x - 2 = 0$ only for $x = 2$ . By part (a) of direct substitution, the numerator approaches $2^3 - 8 = 0$ and the denominator $2 - 2 = 0$ . The limit is of the form $\frac{0}{0}$ , and the quotient law does not apply.
Factor. Let $p(x) = x^3 - 8$ . Since $p(2) = 0$ , part (b) of the factor theorem gives a polynomial $q$ , of degree 2 with leading coefficient 1 (part (a)), such that $p(x) = (x - 2)\, q(x)$ for every real $x$ . Multiplying out gives
$\begin{aligned} &(x - 2)(x^2 + 2x + 4) \\ &\quad = x^3 + 2x^2 + 4x \\ &\qquad - 2x^2 - 4x - 8 \\ &\quad = x^3 - 8 . \end{aligned}$
(11)
So $p(x) = (x - 2)(x^2 + 2x + 4) + 0$ and $p(x) = (x - 2)\, q(x) + 0$ for every real $x$ , and by the uniqueness of the quotient in division of polynomials, dividing $p$ by $x - 2$ , we get $q(x) = x^2 + 2x + 4$ .
Cancel. For $x \ne 2$ the number $x - 2$ is not 0, so
$\begin{aligned} A(x) &= \frac{(x - 2)(x^2 + 2x + 4)}{x - 2} \\ &= x^2 + 2x + 4 . \end{aligned}$
(12)
So $A$ and $g(x) = x^2 + 2x + 4$ are both defined, and equal, at every $x$ with $0 < \abs{x - 2} < 1$ (any $r > 0$ would do).
Compute. $g$ is a polynomial, so by part (a) of direct substitution, $\lim_{x \to 2} g(x) = g(2) = 4 + 4 + 4 = 12$ .
Conclude. By Lemma 1, with $f = A$ , $a = 2$ , $r = 1$ and $L = 12$ , the limit of $A$ is 12.

\boxed{\lim_{x \to 2} \frac{x^3 - 8}{x - 2} = 12}

(13)

So when the side is 2 cm, the volume grows at 12 cubic centimetres per centimetre of side. This is the general fact of the factoring paragraph, with $p(x) = x^3$ and $a = 2$ : the limit is $q(2)$ .

Check. At $x = 2.001$ : $\frac{2.001^3 - 8}{0.001} = \frac{0.012006001}{0.001} = 12.006001$ , as in the table of the figure, and close to 12. ✓ Units: cubic centimetres divided by centimetres. ✓

Rationalising¶

Example 2 (A difference of square roots)

Find $\displaystyle \lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}$ , the limit that a table estimated as 0.50 in an exercise on The Limit of a Function.

The window. Let $f(x) = \frac{\sqrt{1 + x} - 1}{x}$ . If $0 < \abs{x} < 1$ , then $-1 < x < 1$ and $x \ne 0$ , by part (e) of the proposition on distance inequalities, and adding 1 to $-1 < x$ gives $1 + x > 0$ (property 3 of the order rules). So $\sqrt{1 + x}$ is defined (the square-root remark), and $f$ is defined at every $x$ with $0 < \abs{x} < 1$ .
Substitute. By part (a) of direct substitution, $\lim_{x \to 0} (1 + x) = 1$ . Since $1 > 0$ , the root law, part (f) of the limit laws, gives $\lim_{x \to 0} \sqrt{1 + x} = \sqrt{1} = 1$ (since $1 \ge 0$ and $1^2 = 1$ ). By parts (a) and (b), the numerator approaches $1 - 1 = 0$ , and by part (a) the denominator $x$ approaches 0. The limit is of the form $\frac{0}{0}$ .
Rationalise. Let $0 < \abs{x} < 1$ , and $s = \sqrt{1 + x}$ . Then $s \ge 0$ , so $s + 1 \ge 1 > 0$ (adding 1 to both sides, property 3), and we may multiply the numerator and the denominator by $s + 1$ . Since $s^2 = 1 + x$ ,
$\begin{aligned} (s - 1)(s + 1) &= s^2 - 1 \\ &= (1 + x) - 1 \\ &= x . \end{aligned}$
(14)
So, cancelling the non-zero number $x$ ,
$\begin{aligned} f(x) &= \frac{(s - 1)(s + 1)}{x\,(s + 1)} \\ &= \frac{x}{x\,(s + 1)} \\ &= \frac{1}{\sqrt{1 + x} + 1} . \end{aligned}$
(15)
Compute. Let $g(x) = \frac{1}{\sqrt{1 + x} + 1}$ . It is defined at every $x \ge -1$ , since its denominator is at least 1 there, so it agrees with $f$ on the window $0 < \abs{x} < 1$ . By step 2 and part (b) of the limit laws, the denominator approaches $1 + 1 = 2$ , which is not 0; by parts (a) and (d), $\lim_{x \to 0} g(x) = \frac{1}{2}$ .
Conclude. By Lemma 1, with $a = 0$ and $r = 1$ , $\lim_{x \to 0} f(x) = \frac{1}{2}$ .

\boxed{\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x} = \frac{1}{2}}

(16)

The table on The Limit of a Function suggested 0.50; now we have a proof.

Check. At $x = 0.001$ : $g(0.001) = \frac{1}{\sqrt{1.001} + 1} \approx 0.499875$ , the value in that table. ✓

The same form, different answers¶

Example 3 (Three limits of the form

\frac{0}{0}

)

Find each limit, or show that it does not exist: (i) $\displaystyle \lim_{x \to 0} \frac{x^2}{x}$ , (ii) $\displaystyle \lim_{x \to 0} \frac{3x}{x}$ , (iii) $\displaystyle \lim_{x \to 0} \frac{x}{x^2}$ .

All three quotients are defined at every $x \ne 0$ . By part (a) of direct substitution, each numerator and each denominator approaches 0, so all three limits are of the form $\frac{0}{0}$ .

(i) For $x \ne 0$ , cancelling the non-zero number $x$ gives $\frac{x^2}{x} = x$ . Both functions are defined and agree at every $x$ with $0 < \abs{x} < 1$ , and $\lim_{x \to 0} x = 0$ by part (a) of the limit laws. By Lemma 1, the limit is 0.
(ii) For $x \ne 0$ , $\frac{3x}{x} = 3$ . The constant 3 has the limit 3, by part (a) of the limit laws, so by the lemma the limit is 3.
(iii) Let $f(x) = \frac{x}{x^2}$ , and suppose that $\lim_{x \to 0} f(x) = M$ for a real number $M$ . Both $f$ and the function $x$ are defined at every $x \ne 0$ , and $\lim_{x \to 0} x = 0$ , so the product law, part (c) of the limit laws, gives $\lim_{x \to 0} x\,f(x) = 0 \cdot M = 0$ . But for $x \ne 0$ , $x\,f(x) = \frac{x^2}{x^2} = 1$ , so $x\,f(x)$ agrees with the constant 1 at every $x$ with $0 < \abs{x} < 1$ , and by part (a) and the lemma its limit is 1. Two different limits, 0 and 1, contradict uniqueness of limits. So $f$ has no limit at 0.

Conclusion.

(i) $\displaystyle \lim_{x \to 0} \frac{x^2}{x} = 0$ ;
(ii) $\displaystyle \lim_{x \to 0} \frac{3x}{x} = 3$ ;
(iii) $\displaystyle \lim_{x \to 0} \frac{x}{x^2}$ does not exist.

The same form $\frac{0}{0}$ gave 0, 3 and no limit at all. So $\frac{0}{0}$ is not a value; replacing 3 by any real number $c$ in (ii) gives the limit $c$ .

Check. At $x = 0.01$ : (i) $\frac{0.0001}{0.01} = 0.01$ , close to 0 ✓; (ii) $\frac{0.03}{0.01} = 3$ ✓; (iii) $\frac{0.01}{0.0001} = 100$ , and at $x = 0.001$ it is 1000: the values do not settle. ✓

Simplifying a compound fraction¶

Example 4 (Two resistors in parallel, again)

In an example on Limit Laws, two resistors in parallel, of 3 ohms and $t$ ohms, have the combined resistance

R(t) = \frac{3t}{3 + t}

(17)

ohms, for $t > 0$ , and $R(6) = \frac{18}{9} = 2$ . As the second resistance passes 6 ohms, how fast does the combined resistance change? Find the limit of the average rate

f(t) = \frac{R(t) - 2}{t - 6}

(18)

as $t \to 6$ , in ohms per ohm.

The window. $f$ is defined at every $t > 0$ with $t \ne 6$ . If $0 < \abs{t - 6} < 6$ , then $0 < t < 12$ and $t \ne 6$ , by part (e) of the proposition on distance inequalities. So $f$ is defined at every $t$ with $0 < \abs{t - 6} < 6$ .
Substitute. $R$ is the restriction of the rational function $\frac{3t}{3 + t}$ to the open interval $(0, \infty)$ , which contains 6, and the denominator is $9 \ne 0$ at $t = 6$ . By part (c) of direct substitution, $\lim_{t \to 6} R(t) = 2$ , so the numerator approaches $2 - 2 = 0$ (parts (a) and (b) of the limit laws). The denominator approaches 0. The limit is of the form $\frac{0}{0}$ .
One fraction. For $t > 0$ , adding $0 < 3$ and $0 < t$ gives $3 + t > 0$ (property 4 of the order rules), so $3 + t \ne 0$ , and
$\begin{aligned} R(t) - 2 &= \frac{3t - 2(3 + t)}{3 + t} \\ &= \frac{t - 6}{3 + t} . \end{aligned}$
(19)
Cancel. For $t > 0$ with $t \ne 6$ , the number $t - 6$ is not 0, so
$f(t) = \frac{t - 6}{(3 + t)(t - 6)} = \frac{1}{3 + t} .$
(20)
So $f$ and $g(t) = \frac{1}{3 + t}$ are both defined, and equal, at every $t$ with $0 < \abs{t - 6} < 6$ .
Compute. $g$ is a rational function whose denominator is $9 \ne 0$ at $t = 6$ , so by part (b) of direct substitution, $\lim_{t \to 6} g(t) = \frac{1}{9}$ .
Conclude. By Lemma 1, with $a = 6$ and $r = 6$ , $\lim_{t \to 6} f(t) = \frac{1}{9}$ .

\boxed{\lim_{t \to 6} \frac{R(t) - 2}{t - 6} = \frac{1}{9}}

(21)

So near $t = 6$ , a small increase of the second resistance raises the combined resistance by about one ninth of that increase: the rate is $\frac{1}{9}$ ohm per ohm.

Check. At $t = 6.01$ : $R(6.01) = \frac{18.03}{9.01} \approx 2.0011099$ , so $f(6.01) \approx 0.110988$ , and $g(6.01) = \frac{1}{9.01} \approx 0.110988$ , close to $\frac{1}{9} \approx 0.111111$ . ✓ Units: ohms divided by ohms. ✓

Common mistakes¶

Cancelling, then substituting, without the lemma

✗ Wrong: “ $\frac{x^2 - 9}{x - 3} = x + 3$ , so at $x = 3$ the function is 6, and its limit is 6.”

Why: the equality holds only for $x \ne 3$ : the left-hand side is undefined at 3, so it has no value 6 there. Two steps are missing. The functions $\frac{x^2 - 9}{x - 3}$ and $x + 3$ are different functions, and only Lemma 1 says that they have the same limit. And “substituting $x = 3$ ” into $x + 3$ gives the limit only because $x + 3$ is a polynomial (part (a) of direct substitution), not because substituting always works.

✓ Right: for $x \ne 3$ , $\frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} = x + 3$ , so the two functions are defined and agree on the window $0 < \abs{x - 3} < 1$ . By part (a) of direct substitution, $\lim_{x \to 3} (x + 3) = 6$ , and by the lemma $\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = 6$ .

“

\frac{0}{0} = 1

”

✗ Wrong: “In $\lim_{x \to 0} \frac{x^2}{x}$ , the numerator and the denominator both approach 0. A number divided by itself is 1, so $\frac{0}{0} = 1$ , and the limit is 1.”

Why: a number divided by itself is 1 only when it is not 0: $\frac{u}{u} = 1$ needs $u \ne 0$ , because 0 has no reciprocal. $\frac{0}{0}$ is not a number at all. The numerator and the denominator approach 0 at different speeds: $x^2$ is much smaller than $x$ near 0.

✓ Right: the limit is of the form $\frac{0}{0}$ , which says only that the quotient law does not apply. Rewriting gives $\frac{x^2}{x} = x$ for $x \ne 0$ , and the limit is 0 (Example 3). Nor is “ $\frac{0}{0}$ , so the limit does not exist” right: in the same example, $\frac{3x}{x}$ has the limit 3.

Splitting into parts that have no limit

✗ Wrong: “ $\displaystyle \lim_{x \to 3} \frac{\frac{1}{x} - \frac{1}{3}}{x - 3} = \lim_{x \to 3} \frac{1}{x(x - 3)} - \lim_{x \to 3} \frac{1}{3(x - 3)}$ ”, followed by an attempt to compute the two limits on the right.

Why: the difference law, part (b) of the limit laws, needs both limits on the right to exist, as real numbers. Here neither does: the two functions are $\frac{1/x}{x - 3}$ and $\frac{1/3}{x - 3}$ , whose numerators approach $\frac{1}{3}$ , not 0, while their denominators approach 0. Such a quotient has no limit. If $\frac{1/x}{x - 3}$ had a limit $M$ , then, since $x - 3 \to 0$ (parts (a) and (b)), the product law, part (c), would give $(x - 3) \cdot \frac{1/x}{x - 3} \to 0 \cdot M = 0$ . But on the window $0 < \abs{x - 3} < 3$ this product is $\frac{1}{x}$ . Since $\frac{1}{x} \to \frac{1}{3}$ by part (b) of direct substitution, the lemma gives the product the limit $\frac{1}{3}$ , and $0 \ne \frac{1}{3}$ contradicts uniqueness of limits. The same argument applies to $\frac{1/3}{x - 3}$ : there the product is the constant $\frac{1}{3}$ , whose limit is $\frac{1}{3}$ by part (a) of the limit laws (Exercise 11 proves the general fact behind this). Splitting a limit is a conclusion of the laws, never a step you may take before checking their hypotheses.

✓ Right: first bring the numerator to one fraction. For $x \ne 0$ , $\frac{1}{x} - \frac{1}{3} = \frac{3 - x}{3x}$ , so for $x \ne 0$ and $x \ne 3$ the quotient is $\frac{3 - x}{3x(x - 3)} = -\frac{1}{3x}$ . On the window $0 < \abs{x - 3} < 3$ , part (b) of direct substitution and the lemma give the limit $-\frac{1}{9}$ .

Rigorous track¶

Factoring always ends

Let $f = \frac{p}{q}$ be a rational function, where $q$ has degree $m \ge 1$ , and let $a$ be a root of $q$ . By part (b) of the factor theorem, $q(x) = (x - a)\, q_1(x)$ with $q_1$ of degree $m - 1$ (part (a)). If $q_1(a) = 0$ , we factor $q_1$ in the same way, and so on. Each step lowers the degree by one, and a polynomial of degree 0 is a non-zero constant, which is not 0 at $a$ . So after $k$ steps, for some $k$ with $1 \le k \le m$ , we reach a polynomial $Q$ with $Q(a) \ne 0$ such that, for every real $x$ ,

q(x) = (x - a)^k\, Q(x) .

(22)

In the same way, if $p$ is not the zero polynomial, $p(x) = (x - a)^j\, P(x)$ for every real $x$ , with $P(a) \ne 0$ and $j \ge 0$ ( $j = 0$ when $p(a) \ne 0$ ). (If $p$ is the zero polynomial, the limit is 0. By part (b) of the proposition on the domain of a rational function there is an open interval $(\alpha, \beta)$ containing $a$ such that $f$ is defined at every point of it except possibly $a$ ; as in the factoring paragraph, it contains a window $0 < \abs{x - a} < r$ . There $f(x) = \frac{0}{q(x)} = 0$ , so $f$ agrees with the constant 0, whose limit is 0 by part (a) of the limit laws, and the lemma gives $\lim_{x \to a} f(x) = 0$ .) By part (b) of the proposition on the domain of a rational function, applied to $\frac{P}{Q}$ , and as in the factoring paragraph of the main results, there is a window $0 < \abs{x - a} < r$ on which $Q$ has no root; there $x - a \ne 0$ too, so $q(x) = (x - a)^k\, Q(x) \ne 0$ , by part (a) of the sign rules applied $k$ times, and $f$ is defined. On that window:

If $j \ge k$ : cancelling $(x - a)^k$ gives $f(x) = \frac{(x - a)^{j - k} P(x)}{Q(x)}$ , a rational function whose denominator is not 0 at $a$ . By part (b) of direct substitution and Lemma 1, the limit is 0 if $j > k$ , and $\frac{P(a)}{Q(a)}$ if $j = k$ .
If $j < k$ : the limit does not exist. Suppose $\lim_{x \to a} f(x) = L$ . By parts (a), (b) and (e) of the limit laws, $\lim_{x \to a} (x - a)^{k - j} = 0$ , so the product law, part (c), gives $\lim_{x \to a} (x - a)^{k - j} f(x) = 0 \cdot L = 0$ . But on the window, $(x - a)^{k - j} f(x) = \frac{P(x)}{Q(x)}$ , whose limit is $\frac{P(a)}{Q(a)} \ne 0$ , by part (b) of direct substitution and the lemma. This contradicts uniqueness of limits.

So for a rational function, the method of this page always settles a limit of the form $\frac{0}{0}$ : after at most $m$ cancellations it gives the limit, or shows that there is none. What this leaves out: “and so on” stands for an induction on the degree, as in the sketch of the power law on Limit Laws.

When the window is one-sided

The lemma needs a window on both sides of $a$ . Consider $\frac{x - 2}{\sqrt{x - 2}}$ at $a = 2$ . It is defined only for $x > 2$ , and there, with $y = x - 2 > 0$ , it equals $\frac{y}{\sqrt{y}} = \sqrt{y}$ , since $y = \sqrt{y}\,\sqrt{y}$ and $\sqrt{y} \ne 0$ . It is tempting to write

\lim_{x \to 2} \frac{x - 2}{\sqrt{x - 2}} = \lim_{x \to 2} \sqrt{x - 2} = 0 .

(23)

But neither function is defined at any $x < 2$ , so neither is defined on an open interval containing 2 (except at 2), and the definition of a limit applies to neither. The root law agrees: with $L = 0$ , its sign condition asks for $x - 2 \ge 0$ on both sides of 2, and it fails on the left. Rewriting cannot create a window that the function does not have.

Summary¶

A limit is of the form $\frac{0}{0}$ when the numerator and the denominator both approach 0. That is a form, not a value: the quotient law does not apply, and the limit may be any number, or not exist.
The lemma (Lemma 1): if $f(x) = g(x)$ for every $x$ with $0 < \abs{x - a} < r$ , and $\lim_{x \to a} g(x) = L$ , then $\lim_{x \to a} f(x) = L$ . The values at $a$ play no part.
The method: substitute; if the form is $\frac{0}{0}$ , rewrite $f$ for $x \ne a$ as a $g$ to which the laws apply, compute $\lim g$ with the laws, naming each part, and conclude with the lemma.
Factoring: a common root $a$ of numerator and denominator gives a common factor $x - a$ (the factor theorem), which may be cancelled for $x \ne a$ .
Rationalising: $\bigl(\sqrt{u} - \sqrt{v}\bigr)\bigl(\sqrt{u} + \sqrt{v}\bigr) = u - v$ for $u, v \ge 0$ ; then the root law, with $L > 0$ , or $L = 0$ and the sign condition.
Compound fractions: bring them to one fraction, then cancel.

Exercises¶

Exercise 11 (When the denominator approaches zero)

Let $a$ be a real number, and let $f$ be a function such that $\lim_{x \to a} \frac{f(x)}{x - a}$ exists. Prove that $\lim_{x \to a} f(x) = 0$ .

Hint 1

For $x \ne a$ at which $f$ is defined, $f(x) = (x - a) \cdot \frac{f(x)}{x - a}$ . Which law gives the limit of the right-hand side?

Hint 2

You need a window on which $f$ is defined. Where does it come from?

Answer

With $h(x) = \frac{f(x)}{x - a}$ : $f(x) = (x - a)\,h(x)$ for $x \ne a$ , so the product law and the lemma give $\lim_{x \to a} f(x) = 0 \cdot \lim_{x \to a} h(x) = 0$ .

Where this leads¶

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The method of this page, rewrite and then let the lemma carry the limit back, is used again whenever a limit is of the form $\frac{0}{0}$ .