Limits by Direct Substitution

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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) $\lim_{x \to 5} \frac{x - 5}{x^2 - 25}$
2.
Find $\lim_{x \to 2} (3x^2 - x + 4)$ .
3.
Find $\lim_{x \to 4} (x^2 - 3x + 1)$ by direct substitution.
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
5.
Find $\lim_{x \to -1} \frac{2x^2 + 6}{x + 3}$ .
6.
Find $\lim_{x \to 2} \frac{x + 6}{x^2 + 1}$ . Write the answer as a fraction.
7.
Find $\lim_{x \to 5} \frac{x^2 - 25}{x - 5}$ .
8.
Find $\lim_{x \to -3} \frac{x^2 - 9}{x + 3}$ .
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) $5$
10.
Find $\lim_{x \to 4} \frac{x^2 - 2x - 8}{x - 4}$ .
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) $\sqrt{x} + 3$
12.
Find $\lim_{x \to 3} \frac{x^2 - 9}{x^2 - 3x}$ .
13.
Find $\lim_{x \to 1} \frac{x^2 + 2x - 3}{x^2 - 1}$ .
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) $1$
15.
Find $\lim_{x \to 25} \frac{\sqrt{x} - 5}{x - 25}$ by rationalizing.
16.
Find $\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}$ . Give the answer as a fraction or a decimal.
17.
Find $\lim_{x \to 2} \frac{x^3 - 8}{x - 2}$ .
18.
Find $\lim_{x \to 1} \frac{\sqrt{x + 3} - 2}{x - 1}$ .