Question

the police departments machine (from week 2) is set up on a straight stretch of road. at the start of the road, the speed limit is 60 miles per hour. the six graphs from worksheet 1 are graphs the machine recorded. (for the following questions it may help if you figure out what the graph would have looked like for a car that was going the speed limit and draw that in to compare.) 1. did car (a) break the speed limit? what was its velocity, and at what intervals? 2. which of the other cars broke the speed limit? when did the violations happen? (approximate to the nearest.2 seconds and list as intervals.) 3. somewhere near the 4 mile mark, there is a stop sign. did any cars actually stop for it? 4. evaluate the following limits, if they exist. a. $f(x)=\begin{cases}x^{2}+1 & \text{if }x < - 1\sqrt{x + 1}&\text{if }xgeq - 1end{cases}$, $lim_{x
ightarrow - 1^{-}}f(x)$ and $lim_{x
ightarrow - 1^{+}}f(x)$ b. $lim_{t
ightarrow a}\frac{sqrt{3t + 1}-sqrt{3a + 1}}{t - a}$ c. $lim_{t
ightarrow2}\frac{3t^{2}-7t + 2}{2 - t}$

Explanation:

Step1: Analyze left - hand limit of $f(x)$ at $x=-1$

For $x < - 1$, $f(x)=x^{2}+1$. So, $\lim_{x
ightarrow - 1^{-}}f(x)=\lim_{x
ightarrow - 1^{-}}(x^{2}+1)=(-1)^{2}+1 = 2$.

Step2: Analyze right - hand limit of $f(x)$ at $x=-1$

For $x\geq - 1$, $f(x)=\sqrt{x + 1}$. So, $\lim_{x
ightarrow - 1^{+}}f(x)=\lim_{x
ightarrow - 1^{+}}\sqrt{x + 1}=\sqrt{-1 + 1}=0$.

Step3: Solve limit in part b

Multiply numerator and denominator by $\sqrt{3t + 1}+\sqrt{3a+1}$:
\[

$$\begin{align*} \lim_{t ightarrow a}\frac{\sqrt{3t + 1}-\sqrt{3a + 1}}{t - a}&=\lim_{t ightarrow a}\frac{(3t + 1)-(3a + 1)}{(t - a)(\sqrt{3t + 1}+\sqrt{3a+1})}\\ &=\lim_{t ightarrow a}\frac{3(t - a)}{(t - a)(\sqrt{3t + 1}+\sqrt{3a+1})}\\ &=\lim_{t ightarrow a}\frac{3}{\sqrt{3t + 1}+\sqrt{3a+1}}=\frac{3}{2\sqrt{3a + 1}} \end{align*}$$

\]

Step4: Solve limit in part c

Factor the numerator: $3t^{2}-7t + 2=(3t - 1)(t - 2)$.
\[

$$\begin{align*} \lim_{t ightarrow2}\frac{3t^{2}-7t + 2}{2 - t}&=\lim_{t ightarrow2}\frac{(3t - 1)(t - 2)}{-(t - 2)}\\ &=\lim_{t ightarrow2}-(3t - 1)=-(3\times2 - 1)=-5 \end{align*}$$

\]

Answer:

a. $\lim_{x
ightarrow - 1^{-}}f(x)=2$, $\lim_{x
ightarrow - 1^{+}}f(x)=0$
b. $\frac{3}{2\sqrt{3a + 1}}$
c. $- 5$