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Solution to 1.3 Limits-Analytically-(LT3) #481

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103 changes: 92 additions & 11 deletions source/calculus/source/01-LT/02.ptx
Original file line number Diff line number Diff line change
Expand Up @@ -69,6 +69,8 @@
</row>
</tabular>
</table>


<p>
Based on the values of <xref ref="table-limit-numerical1"/>, what is the best approximation for <m>\displaystyle\lim_{x\to 7} f(x)</m>?
<ol marker="A." cols="2">
Expand All @@ -78,6 +80,13 @@
<li><p> the limit is approximately 0.1667</p></li>
<li><p> the limit is approximately 6.9999</p></li>
</ol>

<answer>
<p>
D. The limit is approximately 0.1667
</p>
</answer>

</p>
</activity>

Expand Down Expand Up @@ -156,6 +165,11 @@ the function as <m>x</m> tends to 2?
<li><p> The limit can be approximated to be 1 because the values appear to approach 1 and the graph appears to approach 1, but we should zoom in on the graph to be sure. </p></li>
<li><p> The limit cannot be approximated because the function might not exist at <m>x = 2</m>.</p></li>
</ol>
<answer>
<p>
C. The limit can be approximated to be 1 because the values appear to approach 1 and the graph appears to approach 1, but we should zoom in on the graph to be sure.
</p>
</answer>
</activity>

<activity xml:id = "activity-limits-numerically3">
Expand Down Expand Up @@ -223,6 +237,11 @@ the function as <m>x</m> tends to 2?
<li><p> <m>\displaystyle \lim_{x\rightarrow 1^-}f(x) = 0.5,\,\, \lim_{x\rightarrow 1^+}f(x) = -0.5</m>, and <m>\displaystyle\lim_{x\rightarrow 1}f(x)</m> does not exist</p></li>
<li><p> <m>\displaystyle \lim_{x\rightarrow 1^-}f(x) = 0.5,\,\, \lim_{x\rightarrow 1^+}f(x) = -0.5</m>, and <m>\displaystyle\lim_{x\rightarrow 1}f(x) = 0</m></p></li>
</ol>
<answer>
<p>
B. <m> \displaystyle \lim_{x\to 1^-}f(x) = -0.5,\,\, \lim_{x\rightarrow 1^+}f(x) = 0.5</m>, and <m>\displaystyle\lim_{x\rightarrow 1}f(x)</m> does not exist
</p>
</answer>
</activity>


Expand All @@ -231,6 +250,7 @@ the function as <m>x</m> tends to 2?
<p>Consider the following function <m>f(x)= 3x^3 + 2x^2- 5x+20.</m></p>
</introduction>
<task>
<statement>
<p> Of the following options, at which values given would you evaluate <m>f(x)</m> to best determine <m>\displaystyle \lim_{x \to 2}f(x)</m> numerically?</p>

<ol marker="A." cols="2">
Expand All @@ -239,11 +259,23 @@ the function as <m>x</m> tends to 2?
<li><p> 1.8, 1.9, 2.0, 2.1, 2.2</p></li>
<li><p> 1.0, 1.5, 2.0, 2.5, 3.0</p></li>
</ol>

</statement>
<answer>
<p>
B. 1.98, 1.99, 2.0, 2.01, 2.02
</p>
</answer>
</task>

<task>
<statement>
<p> Use the values that you chose in part (a) to calculate an approximation for <m>\displaystyle\lim_{x \to 2}f(x)</m>. </p>
</statement>
<answer>
<p>
41.22, 41.61, 42, 42.39, 42.78
</p>
</answer>
</task>
<task>
<p> Which value best describes the limit that you obtained in part (b)?</p>
Expand All @@ -254,7 +286,11 @@ the function as <m>x</m> tends to 2?
<li><p> The approximate value is 41.75</p></li>
<li><p> The approximate value is 42</p></li>
</ol>

<answer>
<p>
D. The approximate value is 42.
</p>
</answer>
</task>
</activity>

Expand Down Expand Up @@ -341,13 +377,26 @@ the function as <m>x</m> tends to 2?
<li><p> The limit does not exist because you are dividing by zero when <m>x = 0</m> for <m>f(x).</m></p></li>

</ol>
</statement> </task>
</statement>
<answer>
<p>
C. The limit does not exist because the function is oscillating between -1 and 1.
</p>
</answer>
</task>
<task> <statement>
<p>Would your conclusion that resulted from <xref ref = "activity-limits-numerically5"/> change if the function was <m>f(x) = \cos(1/x)</m> or <m>f(x) = \tan(1/x)</m>?</p>
</statement> </task>
</statement>
<answer>
<p>
Yes, the limit of <m>f(x) = \tan(1/x)</m> would be 0 as <m> x </m> tends to 0 however <m>f(x) = \cos(1/x)</m> oscillates as <m> x </m> tends to 0 .
</p>
</answer>
</task>
</activity>

<activity checkit-seed="0014" checkit-slug="LT2" checkit-title="Limits numerically">

<statement>
<p>
Use technology to complete the following table of values.
Expand All @@ -364,7 +413,9 @@ Then explain how to use it to make an educated guess as to the value of
the limit <me>\displaystyle \lim_{ x\to -3 } \dfrac{ x^{2} - x - 12 }{ x^{2} + 16 \, x + 39 }</me>
</p>
</statement>
<!-- <answer>


<answer>
<p>
<me>
\begin{array}{c|ccccccc}
Expand All @@ -376,7 +427,7 @@ f(x) &amp; -0.71717&amp; -0.7017&amp; -0.70017&amp; DNE&amp; -0.69983&amp; -0.69
<p>
The limit is <m>-0.7</m>.
</p>
</answer> -->
</answer>
</activity>

<aside>
Expand Down Expand Up @@ -466,15 +517,45 @@ The limit is <m>-0.7</m>.
</table>
</introduction>
<task> <statement> <p>What was the average velocity on the first 50 meters? On the second 50 meters? </p>
</statement> </task>
</statement>
<answer>
<p>
The average velocity on the first 50 meters is <m>10.95 </m> meter per second and <m>11.93 </m> meter per second on the second 50 meters.
</p>
</answer>
</task>
<task> <statement> <p>What was the average velocity between 30 and 50 meters? Between 50 and 70 meters? </p>
</statement> </task>
</statement>
<answer>
<p>
The average velocity between 30 and 50 meters is <m>11.62 </m> meter per second and <m>12.19 </m> meter per second between 50 and 70 meters.
</p>
</answer>
</task>
<task> <statement> <p> What was the average velocity between 40 and 50 meters? Between 50 and 60 meters? </p>
</statement> </task>
</statement>
<answer>
<p>
The average velocity between 40 and 50 meters is <m>11.76 </m> meter per second and <m>12.19 </m> meter per second between 50 and 70 meters.
</p>
</answer>
</task>
<task> <statement> <p>What is your best estimate for the Usain's velocity at the instant when he passed the 50 meters mark? This is your estimate for the instantaneous velocity. </p>
</statement> </task>
</statement>
<answer>
<p>
<m>12</m> meter per second.
</p>
</answer>
</task>
<task> <statement> <p>Using the table of values, explain why 50 meters is NOT the best guess for when the instantaneous velocity was the largest. What other point would be more reasonable? </p>
</statement> </task>
</statement>
<answer>
<p>
<m>12</m>
</p>
</answer>
</task>
</activity>

</subsection>
Expand Down
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