S4 Mathematics Practice Book · Unit 7: Limits of polynomial, rational and irrational functions

Unit 7 Limits of polynomial, rational and irrational functions

Key unit competence: Apply limits of polynomial, rational and irrational functions in solving related problems.

Key ideas

$lim_{x \to a} f (x) = L$ means $f (x)$ gets as close as we like to $L$ when $x$ is close enough to $a$ (but $x \neq = a$ ).
For polynomials and for rational functions where the denominator is not 0: substitute, $lim_{x \to a} f (x) = f (a)$ .
One-sided limits: $lim_{x \to a^{-}}$ and $lim_{x \to a^{+}}$ . The limit exists only if both one-sided limits exist and are equal.
Operations: limits of sums, products and quotients are sums, products and quotients of limits (when defined). Squeeze theorem: if $g \leq f \leq h$ and $g, h \to L$ , then $f \to L$ .
Infinite limits: $lim_{x \to a} \frac{k}{x - a}$ is $\pm \infty$ (vertical asymptote $x = a$ ). Limits at infinity of rational functions: compare the highest powers.
Indeterminate forms $\frac{0}{0}$ , $\frac{\infty}{\infty}$ , $\infty - \infty$ , $0 \times \infty$ : remove them by factorising and cancelling, multiplying by the conjugate (for roots), or dividing by the highest power.

Formulas to know

x \to \infty lim \frac{a _{n} x ^{n} + \dots}{b _{m} x ^{m} + \dots} = ⎩ ⎨ ⎧ 0 \frac{a _{n}}{b _{m}} \pm \infty n < m n = m n > m

x \to a lim \frac{x ^{2} - a ^{2}}{x - a} = 2 a

x \to 0 lim \frac{1 + x - 1}{x} = \frac{1}{2}

Common misconceptions

" $\frac{0}{0} = 0$ " or " $\frac{0}{0} = 1$ ". No: $\frac{0}{0}$ is indeterminate; simplify the expression first.
"If $f (a)$ is undefined, the limit at $a$ does not exist." No: $\frac{x ^{2} - 1}{x - 1}$ is undefined at 1 but its limit is 2.

Worked examples

Example 1. Find $lim_{x \to 2} \frac{x ^{2} - 4}{x - 2}$ .

Solution

Form $\frac{0}{0}$ . Factorise: $\frac{( x - 2 ) ( x + 2 )}{x - 2} = x + 2 \to 4$ .

Example 2. Find $lim_{x \to \infty} \frac{3 x ^{2} - x}{2 x ^{2} + 5}$ .

Solution

Divide by $x^{2}$ : $\frac{3 - \frac{1}{x}}{2 + \frac{5}{x ^{2}}} \to \frac{3}{2}$ .

Practice questions

Easier and harder questions are mixed, just like in a real exam.

Find $lim_{x \to - 2} \frac{x ^{2} + 5 x + 6}{x + 2}$ .
Show answer
$\frac{( x + 2 ) ( x + 3 )}{x + 2} = x + 3 \to 1$ .
Common mistake: Answering "undefined" because of $\frac{0}{0}$ .
Find $lim_{x \to 1} \frac{x ^{2} + x - 2}{x ^{2} - 1}$ .
Show answer
$\frac{( x - 1 ) ( x + 2 )}{( x - 1 ) ( x + 1 )} \to \frac{3}{2}$ .
Common mistake: Factorising $x^{2} + x - 2$ as $(x + 1) (x - 2)$ .
Find $lim_{x \to \infty} \frac{4 x ^{3} - 2 x + 7}{2 x ^{3} + x ^{2}}$ and state the horizontal asymptote.
Show answer
2; asymptote $y = 2$ .
Common mistake: Comparing constant terms.
Use the squeeze theorem to find $lim_{x \to \infty} \frac{sin x}{x}$ .
Show answer
$- \frac{1}{x} \leq \frac{sin x}{x} \leq \frac{1}{x}$ and both bounds tend to 0, so the limit is 0.
Common mistake: Thinking the limit does not exist because $sin x$ oscillates.
Find $lim_{x \to 9} \frac{x - 9}{x - 3}$ .
Show answer
$x - 9 = (x - 3) (x + 3)$ , so the limit is $9 + 3 = 6$ .
Common mistake: Multiplying only the denominator by the conjugate.
Find $lim_{x \to \infty} (x^{2} + 4 x - x)$ .
Show answer
$\frac{4 x}{x ^{2} + 4 x + x} \to \frac{4}{2} = 2$ .
Common mistake: Answering $\infty - \infty = 0$ .
Study $lim_{x \to 2^{-}} \frac{1}{x - 2}$ and $lim_{x \to 2^{+}} \frac{1}{x - 2}$ . What does this mean for the graph?
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$- \infty$ and $+ \infty$ : the graph has the vertical asymptote $x = 2$ .
Common mistake: Thinking both sides go to $+ \infty$ .
$f (x) = 2 x - 1$ for $x \leq 3$ , $f (x) = x + k$ for $x > 3$ . Find $k$ so that $lim_{x \to 3} f (x)$ exists.
Show answer
Left limit 5, right limit $3 + k$ : $k = 2$ .
Common mistake: Using only one piece.

Matching and drag-and-drop

Matching

Match each item in column A with the correct answer in column B. Write the letter next to each number.

Column A	Column B
1. Find $lim_{x \to + \infty} \frac{3 x + 1}{x ^{2} + 4}$ .	a. $- \frac{1}{2}$
2. Find $lim_{x \to 1} \frac{x ^{2} - 3 x + 2}{x ^{2} - 1}$ .	b. $0$
3. Find $lim_{x \to 3} \frac{x ^{3} - 27}{x - 3}$ .	c. $27$
4. Find the vertical asymptotes of $f (x) = \frac{x + 1}{x ^{2} - 4}$ .	d. $x = 2$ and $x = - 2$

Show answer

1 → b ( $0$ ) · 2 → a ( $- \frac{1}{2}$ ) · 3 → c ( $27$ ) · 4 → d ( $x = 2$ and $x = - 2$ )

Step 1. Degree of the numerator (1) is less than that of the denominator (2).
Step 2. Divide by $x^{2}$ : $\frac{3/ x + 1/ x ^{2}}{1 + 4/ x ^{2}} \to 0$ .
Step 1. $\frac{( x - 1 ) ( x - 2 )}{( x - 1 ) ( x + 1 )} = \frac{x - 2}{x + 1}$ .
Step 2. At $x = 1$ : $\frac{- 1}{2}$ .
Step 1. $x^{3} - 27 = (x - 3) (x^{2} + 3 x + 9)$ .
Step 2. Cancel: $9 + 9 + 9 = 27$ .
Step 1. The denominator is 0 at $x = \pm 2$ .
Step 2. The numerator is not 0 there ( $3$ and $- 1$ ), so both are vertical asymptotes.

Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
$- \frac{1}{4}$ · $k = - 1$ · $- \frac{1}{2}$ · $8$
1. Find $lim_{x \to 0} \frac{1}{x} (\frac{1}{x + 2} - \frac{1}{2})$ . →
2. $f (x) = x^{2}$ for $x < 1$ and $f (x) = 2 x + k$ for $x \geq 1$ . Find $k$ so that $lim_{x \to 1} f (x)$ exists. →
3. Find $lim_{x \to + \infty} \frac{( 2 x - 1 ) ^{3}}{x ^{3} + 1}$ . →
Show answer
1: $- \frac{1}{4}$ · 2: $k = - 1$ · 3: $8$
- Step 1. $\frac{1}{x + 2} - \frac{1}{2} = \frac{2 - ( x + 2 )}{2 ( x + 2 )} = \frac{- x}{2 ( x + 2 )}$ .
  Step 2. Divide by $x$ : $\frac{- 1}{2 ( x + 2 )} \to - \frac{1}{4}$ .
- Step 1. Left limit: $1^{2} = 1$ . Right limit: $2 + k$ .
  Step 2. $2 + k = 1$ , so $k = - 1$ .
- Step 1. The leading term of $(2 x - 1)^{3}$ is $8 x^{3}$ .
  Step 2. Same degree: $\frac{8}{1} = 8$ .
Matching
Match each item in column A with the correct answer in column B. Write the letter next to each number.
Column A Column B
1. Find $lim_{x \to + \infty} (x^{2} + 4 x - x)$ . a. $2$
2. Find $lim_{x \to 2} \frac{x ^{2} - 4}{x ^{2} + x - 6}$ . b. $1$
3. Find $lim_{x \to 1} \frac{x ^{4} - 1}{x - 1}$ . c. $\frac{4}{5}$
4. Find $lim_{x \to 2} \frac{x ^{2} - 3 x + 2}{x - 2}$ . d. $4$
Show answer
1 → a ( $2$ ) · 2 → c ( $\frac{4}{5}$ ) · 3 → d ( $4$ ) · 4 → b ( $1$ )
- Step 1. Multiply by the conjugate: $\frac{x ^{2} + 4 x - x ^{2}}{x ^{2} + 4 x + x} = \frac{4 x}{x ^{2} + 4 x + x}$ .
  Step 2. Divide by $x$ : $\frac{4}{1 + 4/ x + 1} \to \frac{4}{2} = 2$ .
- Step 1. Substitution gives $\frac{0}{0}$ .
  Step 2. $\frac{( x - 2 ) ( x + 2 )}{( x - 2 ) ( x + 3 )} = \frac{x + 2}{x + 3}$ .
  Step 3. At $x = 2$ : $\frac{4}{5}$ .
- Step 1. $x^{4} - 1 = (x - 1) (x + 1) (x^{2} + 1)$ .
  Step 2. Cancel: $(1 + 1) (1 + 1) = 4$ .
- Step 1. $x^{2} - 3 x + 2 = (x - 1) (x - 2)$ .
  Step 2. Cancel: $x - 1 \to 1$ .
Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
$- \infty$ · $- \frac{1}{2}$ · $0$ · $1$
1. Find $lim_{x \to - \infty} \frac{2 x ^{3} - x}{1 - 4 x ^{3}}$ . →
2. Find $lim_{x \to - 1} \frac{x ^{2} + 3 x + 2}{x + 1}$ . →
3. Find $lim_{x \to - \infty} \frac{x ^{3} + 1}{2 x ^{2} - 1}$ . →
Show answer
1: $- \frac{1}{2}$ · 2: $1$ · 3: $- \infty$
- Step 1. Same degree (3).
  Step 2. Ratio of leading coefficients: $\frac{2}{- 4} = - \frac{1}{2}$ .
- Step 1. $\frac{0}{0}$ : factorise $x^{2} + 3 x + 2 = (x + 1) (x + 2)$ .
  Step 2. Cancel: $x + 2 \to 1$ .
- Step 1. It behaves like $\frac{x ^{3}}{2 x ^{2}} = \frac{x}{2}$ .
  Step 2. As $x \to - \infty$ , $\frac{x}{2} \to - \infty$ .

Column A	Column B
1. Find $lim_{x \to + \infty} (x^{2} + 4 x - x)$ .	a. $2$
2. Find $lim_{x \to 2} \frac{x ^{2} - 4}{x ^{2} + x - 6}$ .	b. $1$
3. Find $lim_{x \to 1} \frac{x ^{4} - 1}{x - 1}$ .	c. $\frac{4}{5}$
4. Find $lim_{x \to 2} \frac{x ^{2} - 3 x + 2}{x - 2}$ .	d. $4$

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