Limits by Direct Substitution

Show your work for each problem.

1.
Which limit can you find right away by direct substitution?
1. a) $\lim_{x \to 1} \frac{x^2 + 3}{x + 1}$
2. b) $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$
3. c) $\lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x}$
4. d) $\lim_{x \to 5} \frac{x - 5}{x^2 - 25}$
Answer: $\lim_{x \to 1} \frac{x^2 + 3}{x + 1}$
At $x = 1$ the denominator $x + 1$ equals $2 \neq 0$ , so $\lim_{x \to 1} \frac{x^2 + 3}{x + 1} = \frac{4}{2} = 2$ . The other three give $\frac{0}{0}$ and must be simplified first.
2.
Find $\lim_{x \to 2} (3x^2 - x + 4)$ .
Answer: 14
$3 \cdot 2^2 - 2 + 4 = 12 - 2 + 4 = 14$ .
3.
Find $\lim_{x \to 4} (x^2 - 3x + 1)$ by direct substitution.
Answer: 5
1. What is $x^2$ at $x = 4$ ? 16
2. What is $-3x$ at $x = 4$ ? -12
3. Add everything: $16 - 12 + 1$ . What is the limit? 5
4.
Direct substitution into $\lim_{x \to 4} \frac{x^2 - 16}{x - 4}$ gives $\frac{0}{0}$ . What should you conclude?
1. a)The limit does not exist
2. b)The limit is 1
3. c)Simplify first, then substitute again
4. d)The limit is 0
Answer: Simplify first, then substitute again
$\frac{0}{0}$ tells you nothing about the limit yet. Factoring gives $\frac{(x - 4)(x + 4)}{x - 4} = x + 4$ , so the limit is $8$ .
5.
Find $\lim_{x \to -1} \frac{2x^2 + 6}{x + 3}$ .
Answer: 4
Numerator: $2 \cdot (-1)^2 + 6 = 8$ . Denominator: $-1 + 3 = 2$ . Limit: $\frac{8}{2} = 4$ .
6.
Find $\lim_{x \to 2} \frac{x + 6}{x^2 + 1}$ . Write the answer as a fraction.
Answer: 8/5
1. What is the denominator $x^2 + 1$ at $x = 2$ ? 5
2. What is the numerator $x + 6$ at $x = 2$ ? 8
3. The denominator is not 0. What is the limit? (fraction) 8/5
7.
Find $\lim_{x \to 5} \frac{x^2 - 25}{x - 5}$ .
Answer: 10
$\frac{(x - 5)(x + 5)}{x - 5} = x + 5$ , and $5 + 5 = 10$ .
8.
Find $\lim_{x \to -3} \frac{x^2 - 9}{x + 3}$ .
Answer: -6
1. What is the numerator $x^2 - 9$ at $x = -3$ ? 0
2. What is the denominator $x + 3$ at $x = -3$ ? 0
3. It is $\frac{0}{0}$ , so factor: $x^2 - 9 = (x + 3)(x - \square)$ . What number goes in the box? 3
4. After cancelling $(x + 3)$ you get $x - 3$ . What is the limit? -6
9.
What is $\lim_{x \to -2} \frac{x^2 + 5x + 6}{x + 2}$ ?
1. a) $1$
2. b) $-1$
3. c) $0$
4. d) $5$
Answer: $1$
Substitution gives $\frac{0}{0}$ . $\frac{(x + 2)(x + 3)}{x + 2} = x + 3$ , and $-2 + 3 = 1$ .
10.
Find $\lim_{x \to 4} \frac{x^2 - 2x - 8}{x - 4}$ .
Answer: 6
1. What is the numerator $x^2 - 2x - 8$ at $x = 4$ ? 0
2. The denominator is also 0. Factor: $x^2 - 2x - 8 = (x - 4)(x + \square)$ . What number goes in the box? 2
3. After cancelling you get $x + 2$ . What is the limit? 6
11.
To evaluate $\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}$ , you multiply the numerator and the denominator by:
1. a) $x + 9$
2. b) $\sqrt{x}$
3. c) $\sqrt{x} - 3$
4. d) $\sqrt{x} + 3$
Answer: $\sqrt{x} + 3$
$(\sqrt{x} - 3)(\sqrt{x} + 3) = x - 9$ , which cancels with the denominator: the limit becomes $\frac{1}{\sqrt{9} + 3} = \frac{1}{6}$ .
12.
Find $\lim_{x \to 3} \frac{x^2 - 9}{x^2 - 3x}$ .
Answer: 2
$\frac{(x - 3)(x + 3)}{x(x - 3)} = \frac{x + 3}{x}$ , and $\frac{3 + 3}{3} = 2$ .
13.
Find $\lim_{x \to 1} \frac{x^2 + 2x - 3}{x^2 - 1}$ .
Answer: 2
1. What is the numerator $x^2 + 2x - 3$ at $x = 1$ ? 0
2. What is the denominator $x^2 - 1$ at $x = 1$ ? 0
3. Cancelling $(x - 1)$ leaves $\frac{x + 3}{x + 1}$ . What is the new numerator $x + 3$ at $x = 1$ ? 4
4. What is the new denominator $x + 1$ at $x = 1$ ? 2
5. What is the limit? 2
14.
What is $\lim_{x \to 1} \frac{x^3 - 1}{x^2 - 1}$ ?
1. a) $\frac{2}{3}$
2. b) $0$
3. c) $\frac{3}{2}$
4. d) $1$
Answer: $\frac{3}{2}$
$\frac{(x - 1)(x^2 + x + 1)}{(x - 1)(x + 1)} = \frac{x^2 + x + 1}{x + 1}$ , and at $x = 1$ : $\frac{3}{2}$ .
15.
Find $\lim_{x \to 25} \frac{\sqrt{x} - 5}{x - 25}$ by rationalizing.
Answer: 1/10
1. What is the numerator $\sqrt{x} - 5$ at $x = 25$ ? 0
2. $(\sqrt{x} - 5)(\sqrt{x} + 5) = x - \square$ . What number goes in the box? 25
3. After cancelling $(x - 25)$ you have $\frac{1}{\sqrt{x} + 5}$ . What is $\sqrt{x} + 5$ at $x = 25$ ? 10
4. What is the limit? (fraction) 1/10
16.
Find $\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}$ . Give the answer as a fraction or a decimal.
Answer: 1/4
After rationalizing: $\frac{x}{x(\sqrt{x + 4} + 2)} = \frac{1}{\sqrt{x + 4} + 2}$ , and at $x = 0$ : $\frac{1}{2 + 2} = \frac{1}{4}$ .
17.
Find $\lim_{x \to 2} \frac{x^3 - 8}{x - 2}$ .
Answer: 12
1. What is the numerator $x^3 - 8$ at $x = 2$ ? 0
2. $x^3 - 8 = (x - 2)(x^2 + 2x + \square)$ . What number goes in the box? 4
3. After cancelling you get $x^2 + 2x + 4$ . What is the limit? 12
18.
Find $\lim_{x \to 1} \frac{\sqrt{x + 3} - 2}{x - 1}$ .
Answer: 1/4
1. What is the numerator $\sqrt{x + 3} - 2$ at $x = 1$ ? 0
2. $(\sqrt{x + 3} - 2)(\sqrt{x + 3} + 2) = (x + 3) - \square$ . What number goes in the box? 4
3. After cancelling $(x - 1)$ you have $\frac{1}{\sqrt{x + 3} + 2}$ . What is $\sqrt{x + 3} + 2$ at $x = 1$ ? 4
4. What is the limit? (fraction) 1/4