Question

please expalin how to do problem 5 in exercise 5 in matlab

Unfortunately, not all graphs will immediately give us an answer. The following exercise illustrates a possible trouble identifying a limit from the graph and then shows how a theorem from calculus can help us figure out the actual limit. Exercise 5 We wish to find the limit of the oscillating function as x approaches0 Plot the function f(x) x sin(1/x), over the interval [-n,n] using 100 points for x 1. Try to zoom in on the limit. What happens near x 0 that prevents you from getting the correct answer? 2. Graph the same function using 1000 points on top of your current graph (hold) using a different color 3. The graph with 100 points didn't allow us to zoom in far enough to see the limit as this function oscillates so often that when we zoom in we eventually get just straight lines Do you have this same problem using 1000 points? Select exactly one of the choices O Yes No 4. Motivated by the squeeze theorem, on the same graph (hold on) plot both y-xl and y-xl (xl is abs (x) in MATLAB) What relationship, if any, can be observed between f(x)x sin(1/x) and these two lines? Select exactly one of the choices O Ixl 2 f(x) and -Ixl s f(x) 5. Graphically estimate the limit as x-0 What is the limit? (If it does not exist type 'DNE')? Enter a number

Answer 1

MATLAB Code:

close all
clear
clc

% Part 1
x = linspace(-0.2,0.2,100);
f = x.*sin(1./x);
plot(x,f), hold on
disp('With 100 points, we observe the same value from -0.002 to 0.002.')

% Part 2
x = linspace(-0.2,0.2,1000);
f = x.*sin(1./x);
plot(x,f), hold on

% Part 3
disp('Even with 1000 points, we observe the same value (f = 0) around x = 0, which may indeed be the limiting value.')

% Part 4 & 5
y1 = abs(x);
y2 = -abs(x);
plot(x,y1,x,y2)
legend('100 Points', '1000 Points', 'y = |x|', 'y = -|x|')
xlabel('x')
disp('The plot of f(x) with 1000 points is between the plots of |x| and -|x|, which suggests that the limit is zero.')
disp('For Part 4, Option (3)')
disp('For Part 5, Limit = 0')

Plot:

100 Points Pints 0.15 0.1 0.05 ง.05 -0.1 0.15 0.1 0.1 0.15 02

Zoomed Plot:

103 100 Points Pints x 10ง

Output:

With 100 points, we observe the same value from -0.002 to 0.002.
Even with 1000 points, we observe the same value (f = 0) around x = 0, which may indeed be the limiting value.
The plot of f(x) with 1000 points is between the plots of |x| and -|x|, which suggests that the limit is zero.
For Part 4, Option (3)
For Part 5, Limit = 0