The objective of this lab is to enhance understanding of limit evaluation techniques, particularly focusing on factoring, multiplying by a smart one, and using symbolic computation tools such as Maxima to solve problems involving limits. Students will:
By the end of this lab, students will have a solid grasp of various limit evaluation techniques, proficiency in using Maxima for symbolic computation, and improved problem-solving skills for limits and related concepts.
Using the formatting of reporting learned on the last lab, follow the explanations, and complete all the questions (Q:) on each section.
as if we solved it by hand:
\[ \lim_{x\rightarrow 2} \frac{x^2-6x+8}{x^2-4} \]
(%i1) define(y(x),(x^2-6*x+8)/(x^2-4));(%o1) y(x):=(x^2-6*x+8)/(x^2-4)(%i2) 'limit(y(x),x,2);(%o2) 'limit((x^2-6*x+8)/(x^2-4),x,2)(%i3) numFactor: factor(x^2-6*x+8);(%o3) (x-4)*(x-2)(%i4) denFactor: factor(x^2-4);(%o4) (x-2)*(x+2)(%i5) 'limit(ratsimp(y(x)),x,2);(%o5) 'limit((x-4)/(x+2),x,2)(%i6) limit(ratsimp(y(x)),x,2);(%o6) -(1/2)Q: What the rastsimp() function did? why is good for us do that?(%i7) plot2d(y(x),[x,1,3]);
Q: Observing the plot, what you can comment about the limit?\[ \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{x-1} \]
(%i8) define(y(x),(sqrt(x)-1)/(x-1));(%o8) y(x):=(sqrt(x)-1)/(x-1)(%i9) 'limit(y(x),x,1);(%o9) 'limit((sqrt(x)-1)/(x-1),x,1)\[= \lim_{x\rightarrow 1} \left(\frac{\sqrt{x}-1}{x-1}\right) \cdot \left(\frac{\sqrt{x}+1}{\sqrt{x}+1}\right) \] \[= \lim_{x\rightarrow 1} \frac{(x-1)}{(x-1)\sqrt{x}+1}\]
Q: How multiplying by a smart one helps?(%i10) radcan(y(x));(%o10) 1/(sqrt(x)+1)Q: What the radcan() function did?Maxima is strong! can calculate the limits for you very easily1:
(%i11) limit(y(x),x,1);(%o11) 1/2(%i12) plot2d(y(x),[x,0,2]);
Q: Observing the plot, what you can comment about the limit?If we calculate the slope of the secant line as the two points get closer to each other, we understand that converge to the slope of the tangent line for a single point.
\[\lim_{\Delta x \rightarrow 0} m_{sec} = m_{tan}\]
Now, if we have an exponential of base 2, \(2^x\), and we calculate the convergence of the secant line to the tangent line at \(x=0\) we observe that:
\[\boxed{f(x)=2^x}\]
/* Define the function f(x) */
f(x) := 2^x;
/* Define the msec function */
define(msec(x1, x2), (f(x2) - f(x1)) / (x2 - x1));
/* List of specific values */
val: [2, 1, 0.5, 0.1, 0.01, 0.001];
/* Loop to evaluate msec(0, i) for each value in the list */
for i in val do (
display(msec(0, i))
);Q: After running this script, explain what it does. As the two points get closer what happen with the slope of the secant?\[\boxed{f(x)=3^x}\]
/* Define the function f(x) */
f(x) := 3^x;
/* Define the msec function */
define(msec(x1, x2), (f(x2) - f(x1)) / (x2 - x1));
/* List of specific values */
val: [2, 1, 0.5, 0.1, 0.01, 0.001];
/* Loop to evaluate msec(0, i) for each value in the list */
for i in val do (
display(msec(0, i))
);Q: After running this script, explain what it does. As the two points get closer what happen with the slope of the secant?\[\boxed{f(x)=4^x}\]
/* Define the function f(x) */
f(x) := 4^x;
/* Define the msec function */
define(msec(x1, x2), (f(x2) - f(x1)) / (x2 - x1));
/* List of specific values */
val: [2, 1, 0.5, 0.1, 0.01, 0.001];
/* Loop to evaluate msec(0, i) for each value in the list */
for i in val do (
display(msec(0, i))
);Q: After running this script, explain what it does. As the two points get closer what happen with the slope of the secant?Q: Using maxima calculate \(\log(2)\), \(\log(3)\), and \(\log(4)\)(%i13) log(2),numer;(%o13) 0.6931471805599453(%i14) log(3),numer;(%o14) 1.0986122886681098(%i15) log(4),numer;(%o15) 1.3862943611198906Q: How this 3 quantities relate with the slopes of the secant we calculated before?Q: What you can conclude about the slopes of any exponential at \(x=0\)?Q: What is the definition of \(e\) ?Assume the functions \(f,g,\text{ and }h\) satisfy \(f(x)\leq g(x) \leq h(x)\) for all values of \(x\) near \(a\), except possibly at \(a\).
If \[\lim_{x\rightarrow a} f(x)=\lim_{x\rightarrow a} h(x)=L\], then \[\lim_{x\rightarrow a} g(x) =L\]
\[ f(x) = x^2,\qquad g(x) =x^2\sin \frac{1}{x}, \qquad h(x) = -x^2 \]
(%i16) plot2d([x^2,x^2*sin(1/x),-x^2],[x,-.1,.1]);
Q: What you can conclude about the limit of \(g(x)\) as \(x\) goes to zero?Answer here:
Q: Compute using maxima the limit of g(x) as \(x\) goes to zeroAnswer here:
Q: Calculate using maxima:Q:Explain how you will solve the same problem by handAnswer here:
Cell -> Insert Image)Shift+Enter in each code chunk.Ctrl+RFINAL_Lab2_Calc1_YOURLASTNAME_mmddyy.wxmx.)