Infinite Limits

Show your work for each problem.

1.
Find $\lim_{x \to 0} \frac{1}{x^2}$ .
1. a) $-\infty$
2. b) $0$
3. c) $1$
4. d) $\infty$
Answer: $\infty$
For $x = \pm 0.1$ we get $\frac{1}{0.01} = 100$ , for $x = \pm 0.01$ we get $10000$ . The values grow without bound on both sides, so $\lim_{x \to 0} \frac{1}{x^2} = \infty$ .
2.
The function $f(x) = \frac{7}{x-6}$ has a vertical asymptote $x = a$ . What is $a$ ?
Answer: 6
$x - 6 = 0$ gives $x = 6$ , and the numerator is $7 \neq 0$ . So the function blows up near $x = 6$ and $a = 6$ .
3.
Find $\lim_{x \to 5^+} \frac{1}{x-5}$ .
Answer: $\infty$
1. What is $x - 5$ for $x = 5.01$ ? 0.01
2. What is $\frac{1}{x-5}$ for $x = 5.01$ ? 100
3. What is the right-hand limit? $\infty$
4.
Where is the vertical asymptote of $f(x) = \frac{3}{x+5}$ ?
1. a) $x = 5$
2. b) $y = 0$
3. c) $x = 3$
4. d) $x = -5$
Answer: $x = -5$
The denominator $x + 5$ is $0$ at $x = -5$ , and the numerator $3$ is never $0$ . So $x = -5$ is the vertical asymptote. ( $y = 0$ is a horizontal asymptote.)
5.
For $f(x) = \frac{1}{(x-2)^2}$ , compute $f(2.01)$ .
Answer: 10000
$(0.01)^2 = 0.0001$ and $\frac{1}{0.0001} = 10000$ . Very close to $x = 2$ the function is already huge, a sign of the infinite limit there.
6.
Find $\lim_{x \to 0} \frac{3}{x^2}$ .
Answer: $\infty$
1. What is $x^2$ for $x = 0.1$ (and also for $x = -0.1$ )? 0.01
2. What is $\frac{3}{x^2}$ for $x = 0.1$ and $x = -0.1$ ? 300
3. What is the limit? $\infty$
7.
Find the vertical asymptote of $g(x) = \frac{x+4}{x-1}$ .
Answer: $x = 1$
1. For which $x$ is the denominator $0$ ? 1
2. What is the numerator at that $x$ ? 5
3. Which line is the vertical asymptote? $x = 1$
8.
Find $\lim_{x \to 4^-} \frac{2}{x-4}$ .
1. a) $-\infty$
2. b) $0$
3. c) $\infty$
4. d) $2$
Answer: $-\infty$
For $x < 4$ the denominator $x - 4$ is a small negative number (for example $3.99 - 4 = -0.01$ ), and the numerator $2$ is positive. Positive over small negative gives $\frac{2}{-0.01} = -200$ and lower, so the limit is $-\infty$ .
9.
For $f(x) = \frac{1}{x-2}$ , compute $f(1.999)$ .
Answer: -1000
$\frac{1}{-0.001} = -1000$ . Just left of $x = 2$ the values are large and negative, which matches $\lim_{x \to 2^-} \frac{1}{x-2} = -\infty$ .
10.
Find $\lim_{x \to 1^-} \frac{x+3}{x-1}$ .
Answer: $-\infty$
1. What does the numerator $x + 3$ approach as $x \to 1$ ? 4
2. What is the denominator $x - 1$ for $x = 0.99$ ? -0.01
3. What is $\frac{x+3}{x-1}$ for $x = 0.99$ ? -399
4. What is the left-hand limit? $-\infty$
11.
What is true about $\lim_{x \to 1} \frac{1}{x-1}$ ?
1. a) $\infty$
2. b) $-\infty$
3. c)It does not exist
4. d) $1$
Answer: It does not exist
Left of $1$ : $\frac{1}{0.9 - 1} = -10$ , the values go to $-\infty$ . Right of $1$ : $\frac{1}{1.1 - 1} = 10$ , the values go to $\infty$ . The two sides disagree, so the two-sided limit does not exist and cannot be written as $\infty$ or $-\infty$ .
12.
Find the one-sided limits and the two-sided limit of $f(x) = \frac{x}{x-1}$ at $x = 1$ .
Answer: $\lim_{x \to 1^-} f(x) = -\infty,\ \lim_{x \to 1^+} f(x) = \infty$
1. What is $f(0.9)$ ? -9
2. What is $f(1.1)$ ? 11
3. What is $\lim_{x \to 1^-} f(x)$ ? $-\infty$
4. What is $\lim_{x \to 1^+} f(x)$ ? $\infty$
5. What is $\lim_{x \to 1} f(x)$ ? does not exist
13.
How many vertical asymptotes does $f(x) = \frac{x+1}{x^2-4x+3}$ have?
Answer: 2
$x^2 - 4x + 3 = (x-1)(x-3)$ , so the denominator is $0$ at $x = 1$ and $x = 3$ . The numerator there is $2$ and $4$ , both not $0$ . So there are $2$ vertical asymptotes: $x = 1$ and $x = 3$ .
14.
Find $\lim_{x \to -2} \frac{x-3}{(x+2)^2}$ .
Answer: $-\infty$
1. What does the numerator $x - 3$ approach as $x \to -2$ ? -5
2. What is $\frac{x-3}{(x+2)^2}$ for $x = -1.9$ ? -490
3. What is $\frac{x-3}{(x+2)^2}$ for $x = -2.1$ ? -510
4. What is the limit? $-\infty$
15.
Which function has a vertical asymptote at $x = 3$ ?
1. a) $\frac{1}{x+3}$
2. b) $\frac{x-3}{x^2-9}$
3. c) $\frac{x+3}{x-3}$
4. d) $\frac{x^2-9}{x-3}$
Answer: $\frac{x+3}{x-3}$
$\frac{x+3}{x-3}$ : denominator $0$ at $x = 3$ , numerator $6 \neq 0$ , so it blows up there. $\frac{x-3}{x^2-9} = \frac{1}{x+3}$ has a hole at $x = 3$ (its asymptote is $x = -3$ ), $\frac{x^2-9}{x-3} = x + 3$ only has a hole at $x = 3$ , and $\frac{1}{x+3}$ has its asymptote at $x = -3$ .
16.
The function $f(x) = \frac{x^2-1}{x^2+2x-3}$ has exactly one vertical asymptote $x = a$ . What is $a$ ?
Answer: -3
$\frac{(x-1)(x+1)}{(x-1)(x+3)}$ . At $x = 1$ both are $0$ : after canceling, $\frac{x+1}{x+3} \to \frac{2}{4} = \frac{1}{2}$ , a hole. At $x = -3$ the numerator is $9 - 1 = 8 \neq 0$ , so $a = -3$ .
17.
Analyze $f(x) = \frac{x^2-4}{x^2-5x+6}$ at the zeros of its denominator.
Answer: $x = 3$
1. What is the smaller zero of the denominator? 2
2. What is the larger zero of the denominator? 3
3. The numerator is also $0$ at the smaller zero. Cancel the common factor: what is $\lim_{x \to 2} f(x)$ ? -4
4. At which $x$ is the vertical asymptote? 3
5. What is $\lim_{x \to 3^+} f(x)$ ? $\infty$
18.
Find $\lim_{x \to 2^+} \frac{x^2+1}{4-x^2}$ .
Answer: $-\infty$
1. What does the numerator $x^2 + 1$ approach as $x \to 2$ ? 5
2. What is the denominator $4 - x^2$ at $x = 2$ ? 0
3. What is $4 - x^2$ for $x = 2.01$ ? -0.0401
4. What is the right-hand limit? $-\infty$