AP Calculus AB Unit 1 FRQ Practice: Limits and Continuity

Solution

1. Work with the difference $D(t)=N(t)-S(t)$; it is continuous because $N$ and $S$ are.
2. On the calculator, $S(0)=73.000$, $S(2)=62.574$, $S(3)=55.875$, $S(7)=47.973$, $S(11)=58.225$; subtracting these from the table values gives $D(0)=2.000$, $D(2)=-1.574$, $D(3)=-2.875$, $D(7)=-2.973$, $D(11)=0.775$.
3. $D$ changes sign 2 times between consecutive table times, on $0<t<2$ and $7<t<11$; the IVT gives a zero of $D$ in each.

Solution

1. $e^{-t\mathrm{/}8}\to 0$ as $t\to \mathrm{\infty }$ and $\cos \text{\hspace{0.17em}⁣}\left({\displaystyle \frac{t}{2}}\right)$ stays between $-1$ and $1$, so the product $18e^{-t\mathrm{/}8}\cos \text{\hspace{0.17em}⁣}\left({\displaystyle \frac{t}{2}}\right)$ is squeezed to $0$ and $S(t)\to 55$.

Model answer. The end behavior is described by $\underset{t\to \mathrm{\infty }}{\lim }S(t)$. Since $-1\le \cos \text{\hspace{0.17em}⁣}\left({\displaystyle \frac{t}{2}}\right)\le 1$, $-18e^{-t\mathrm{/}8}\le S(t)-55\le 18e^{-t\mathrm{/}8}$ for $t\ge 0$. Both bounds approach $0$ as $t\to \mathrm{\infty }$, so by the squeeze theorem $\underset{t\to \mathrm{\infty }}{\lim }S(t)=55$: according to the model, the rate at which the ticket office fills orders approaches $55$ orders per hour in the long run.

Solution

1. Graph $y=S(t)$ and $y=57$ on $0\le t\le 11$ (or find the zeros of $S(t)-57$).
2. The graphs cross twice in $0<t<11$: at $t=2.824$, where $S$ falls below $57$, and at $t=10.248$, where it rises back above.

Solution

1. The formula piece is continuous for $t>-2$, so its limit from the right at $t=10$ is its value there, ${\displaystyle \frac{670+k}{12}}$.
2. Continuity at $t=10$ needs this to equal the table value $F(10)=46$: $670+k=46\cdot 12=552$.
3. So $k=552-670=-118$.

Solution

1. Just before $t=34$, $F(t)={\displaystyle \frac{67t-118}{t+2}}$, which is continuous there, so the limit from the left is its value, $60$.
2. Just after $t=34$, $F(t)={\displaystyle \frac{67t-118}{t+2}}+6$, so the limit from the right is $60+6=66$.
3. $60\mathrm{\ne }66$: two finite one-sided limits that differ make a jump; a removable discontinuity would need them equal.

Model answer. With $k=-118$, $\underset{t\to 34^{-}}{\lim }F(t)=$${\displaystyle \frac{67(34)-118}{34+2}}=$${\displaystyle \frac{2160}{36}}=$$60$ and $\underset{t\to 34^{+}}{\lim }F(t)=60+6=66$. The discontinuity is a jump: both one-sided limits are finite, but they are not equal, so $\underset{t\to 34}{\lim }F(t)$ does not exist and no value of $F(34)$ could make $F$ continuous there.

Solution

1. Only the formula for $t>34$ matters as $t\to \mathrm{\infty }$: the fraction has numerator and denominator of the same degree, so it approaches the ratio of the leading coefficients, $67$.
2. Adding the extra $6$ people per minute gives the long-run rate $73$ people per minute.

Model answer. The long-run behavior is described by $\underset{t\to \mathrm{\infty }}{\lim }F(t)$. For $t>34$, $F(t)={\displaystyle \frac{67t-118}{t+2}}+6$; dividing the numerator and the denominator of the fraction by $t$ gives ${\displaystyle \frac{67-{\displaystyle \frac{118}{t}}}{1+{\displaystyle \frac{2}{t}}}}$, which approaches $67$. So $\underset{t\to \mathrm{\infty }}{\lim }F(t)=67+6=73$: in the long run, the rate at which fans enter the stadium approaches $73$ people per minute.

Solution

1. From both sides the graph of $f$ approaches $(2,-2)$; the value $f(2)=0$ does not affect the limit.
2. $g$ is a polynomial, so its limit is its value, $-9$. The limit of a product is the product of the limits: $18$.

Solution

1. At $x=7$ the graph has a corner but no break: both one-sided limits and the value are $4$.
2. $f$ jumps at $x=5$, but $g(5)=0$: the product of a bounded function and one that approaches $0$ approaches $0$ from both sides, and the value of the product there is $0$ too.

Model answer. $f$ is continuous at $x=7$ because $\underset{x\to 7}{\lim }f(x)=f(7)=4$. $f(x)g(x)$ is continuous at $x=5$: $g(5)=0$ and $f$ is bounded near $x=5$, so $\underset{x\to 5^{-}}{\lim }f(x)g(x)=1\cdot 0=0$ and $\underset{x\to 5^{+}}{\lim }f(x)g(x)=3\cdot 0=0$, which equal the value $f(5)g(5)=1\cdot 0=0$.

Solution

1. $g\left({\displaystyle \frac{20}{3}}\right)=5$, the $x$-value of the jump of $f$, and $g$ increases through $5$ there.
2. So as $x$ approaches ${\displaystyle \frac{20}{3}}$ from the left, the inputs $g(x)$ approach $5$ from the left, and as $x$ approaches ${\displaystyle \frac{20}{3}}$ from the right, from the right; $f$ approaches $1$ and $3$ along them (part (a)).
3. $1\mathrm{\ne }3$, so the two-sided limit of $f\left(g(x)\right)$ does not exist.

Solution

1. At $x=0$ and just to its right, $f(x)=-2\cos x+8$, which is continuous, so the limit from the right and the value are both $-2+8=6$.
2. $f$ is continuous at $x=0$ because $\underset{x\to 0^{-}}{\lim }f(x)=$$\underset{x\to 0^{+}}{\lim }f(x)=$$f(0)=$$6$.

Solution

1. Just left of $x=\pi$, $f(x)=-2\cos x+8$, which is continuous: its limit is its value, $-2\cos \pi +8=10$.
2. Just right of $x=\pi$, the right piece is continuous at $x=\pi$ (its denominator is $\pi$ there), and $\sin \pi =0$, so its limit is $13$.

Solution

1. The sine factors oscillate without a limit, but they stay between $-1$ and $1$; divided by $x$, they are trapped between bounds that approach $0$.
2. So the left piece is squeezed to $0$ as $x\to -\mathrm{\infty }$, and the right piece, $13$ plus a term squeezed to $0$, approaches $13$ as $x\to \mathrm{\infty }$.

Model answer. For $x<0$, $\mathrm{\mid }\sin (3x)\mathrm{\mid }\le 1$, so $-{\displaystyle \frac{2}{\mathrm{\mid }x\mathrm{\mid }}}\le {\displaystyle \frac{2\sin (3x)}{x}}\le {\displaystyle \frac{2}{\mathrm{\mid }x\mathrm{\mid }}}$. Both bounds approach $0$ as $x\to -\mathrm{\infty }$, so by the squeeze theorem $\underset{x\to -\mathrm{\infty }}{\lim }f(x)=0$. For $x>\pi$, $f(x)=13-{\displaystyle \frac{\sin x}{x}}$ and $\mathrm{\mid }\sin x\mathrm{\mid }\le 1$, so $13-{\displaystyle \frac{1}{x}}\le f(x)\le 13+{\displaystyle \frac{1}{x}}$. Both bounds approach $13$ as $x\to \mathrm{\infty }$, so $\underset{x\to \mathrm{\infty }}{\lim }f(x)=13$.