Unit 1 practice test: Limits and Continuity

On both sides of $x = -2$ the graph rises without bound, so $\lim_{x \to -2} g(x) = \infty$.

• $\lim_{x \to -2} g(x) = -\infty$: The graph goes up, not down, near $-2$.
• $\lim_{x \to -2^-} g(x)$$= -\infty$ and $\lim_{x \to -2^+} g(x) = \infty$: The left side also rises near $-2$.
• $\lim_{x \to -2} g(x) = 0$: Near $-2$ the values are huge, not near 0.
• Correct: $\lim_{x \to -2} g(x) = \infty$: Both sides go up: the two-sided behavior is $\infty$.

Left of 1 the graph plunges down; right of 1 it shoots up. Each one-sided limit is infinite, with opposite signs.

• Correct: $\lim_{x \to 1^-} g(x) = -\infty$ and $\lim_{x \to 1^+} g(x) = \infty$: Down on the left, up on the right.
• $\lim_{x \to 1^-} g(x) = \infty$ and $\lim_{x \to 1^+} g(x) = -\infty$: The directions are reversed.
• $\lim_{x \to 1} g(x) = \infty$ from both sides of $x = 1$: The left side heads down, not up.
• $\lim_{x \to 1} g(x) = 0$, since the graph crosses the axis: Near 1 the graph is unbounded; crossing the axis elsewhere does not matter.

$\infty$ is not a number. Writing $= \infty$ is a precise way of saying how the limit fails to exist: the values grow beyond every bound.

• $g(-2)$ equals $\infty$, a very large number that the graph reaches: $g(-2)$ is undefined, and $\infty$ is not a number a function can equal.
• Correct: $g(x)$ grows without bound near $-2$; the limit is not a number: An infinite limit is a kind of limit that does not exist, described by its direction.
• The limit exists and equals the largest value of $g$ on the graph: $g$ has no largest value near $-2$; that is the point.
• $g$ approaches the horizontal line $y = \infty$ as $x$ nears $-2$: There is no line $y = \infty$; the graph runs along the vertical asymptote $x = -2$.

The conjugate of $\sqrt{x + 8} - 3$ is $\sqrt{x + 8} + 3$; their product is $(x + 8) - 9 = x - 1$, which cancels with the bottom.

• Correct: $\sqrt{x + 8} + 3$: The conjugate flips the middle sign.
• $\sqrt{x + 8} - 3$: Multiplying by the same expression gives $(\sqrt{x + 8} - 3)^2$, which keeps the square root.
• $x + 1$: $x + 1$ does not remove the square root.
• $\sqrt{x + 8}$: This gives $x + 8 - 3\sqrt{x + 8}$, which still has a square root.

Left values (0.167132, 0.166713, 0.166671) fall toward $0.1666\ldots$ and right values (0.166662, 0.166620, 0.166206) rise toward it. The algebra gives exactly $\frac{1}{6}$: two representations of the same limit.

• Correct: They agree: both sides close in on $\frac{1}{6} = 0.1666\ldots$: Both sides head to $\frac{1}{6}$.
• They disagree: the table suggests 0.17, the largest value shown.: The trend matters, not one value: the values keep closing in on $0.1666\ldots$.
• The table shows the limit does not exist, because the values differ on each side.: Values on each side differ slightly but head to the same number.
• The table shows $g(1) = \frac{1}{6}$.: The table does not include $x = 1$, where $g$ is undefined.

At $x = 1$ the top is $\sqrt{9} - 3 = 0$ and the bottom is 0, so the formula has no value there. A limit does not need one.

• Correct: $g(1)$ is not defined: substituting gives $\frac{0}{0}$.: Zero over zero has no value.
• $g(1) = 0$, which is not interesting.: The formula does not give 0 at $x = 1$; it gives no value at all.
• $g(1)$ equals the limit, so the table does not need to show it.: $g(1)$ does not exist, so it cannot equal the limit.
• Tables of values skip whole-number inputs.: Any input near the target can appear; the target itself is skipped.