CoCalc Public FilesScalese Looking At Limits.sagews
Authors: Laura Gross, TJ Scalese
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Compute Environment: Ubuntu 20.04 (Default)
#Thomas Scalese #Learning To Look At Limits # Question 1. Find the limit of the function F(x) = x^2 + 3*x + 2, as x approaches 3 from the left and right side of the graph. Then determine what the limit of f(x) is when x approaches 3 from both sides. Then plot the function, and use the correct paramters, to visualize the limit as f(x) approaches 3. f(x) = x^2 + 3*x + 2 N(f(2.9)) N(f(2.99)) N(f(2.999)) #as x approaches 3 from the left side, f(x) approaches 20 N(f(3.1)) N(f(3.01)) N(f(3.001)) #as x approaches 3 from the right side, f(x) approaches 20 #Becasue as x approaches 3 from the left side, f(x) approaches 20, and as x approaches 3 from the right side, f(x) also approaches 20, we can determine that the limit of f(x) as x approaches 3 from both sides is 20. We can also solve this mathematicaly by plugging in 3 for x in f(x) N(f(3)) plot(x^2 + 3*x +2, 0,4) + plot(20, 0,4, color="red")
19.1100000000000 19.9101000000000 19.9910010000000 20.9100000000000 20.0901000000000 20.0090010000000 20.0000000000000
# Question 2. Find the limit of the funcation, f(x) = (x-4)/(x-2), as x approaches 2 from the left and right side. Then determine what the limit of f(x) as x approaches 2 from both sides is. If the limit does not exist please state what type of discontinuity is. f(x) = (x-4)/(x-2) N(f(1.9)) N(f(1.99)) N(f(1.999)) N(f(1.9999)) N(f(1.99999)) N(f(1.999999)) #as x approaches 2 from the left side, f(x) approaches positive infinity N(f(2.1)) N(f(2.01)) N(f(2.001)) N(f(2.0001)) N(f(2.00001)) N(f(2.000001)) #as x approaches 2 from the right side, f(x) approaches negative infinity plot((x-4)/(x-2), 0,4) #Because the limits of f(x) as x approaches 2 from the left and right side are postive infinity and negative infinity respectively you can say that the limit of f(x), when x approaches 2, does not exist. You can also see this by plotting the graph, the graph will never touch x=2. Becasue of this the funcation has an infinite disontinuity.
21.0000000000000 201.000000000000 2001.00000000022 20001.0000000022 200000.999998690 2.00000100016453e6 -19.0000000000000 -199.000000000004 -1999.00000000022 -19998.9999999578 -199998.999998690 -1.99999899972044e6
#Question 3. Find the limit of the funcation, f(x) = (x)/(abs(x)), as x approaches 0 from the left and right sides of the graph. Then with that information determine what the limit of f(x) when x approaches 0 from both sides is. If the limit does not exist please determine what type of disontinuity it is. f(x) = (x)/(abs(x)) N(f(-0.1)) N(f(-0.01)) N(f(-0.001)) #as x approaches 0 from the left side, f(x) approaches -1 N(f(0.1)) N(f(0.01)) N(f(0.001)) #as x approaches 0 from the right side, f(x) approaches 1 plot((x)/(abs(x))) #with this information you can determine that the limit of f(x) as x appraoches 0 from both sides, does not exist. You can also see this by graphing f(x). On the graph you can see that f(x) will never touch as one sigle point at x=0. This is considered a jump discontinuity.
-1.00000000000000 -1.00000000000000 -1.00000000000000 1.00000000000000 1.00000000000000 1.00000000000000
#Question 4. Find the limit of the funcation, f(x) = (x^2 - 2*x -3)/(x+1), as x approaches -1 from the left and right sides of the graph. Then determine what f(x)=-1 is. Determine if there is a removable discontinuity. f(x) = (x^2 - 2*x -3)/(x+1) N(f(-0.9)) N(f(-0.99)) N(f(-0.999)) #as x apporaches -1 from the left side, f(x) approaches -4 N(f(-1.1)) N(f(-1.01)) N(f(-1.001)) #as x approaches -1 from the right side, f(x) appraoches -4 #With this function f(x) you are unable to plug in x=-1. This is because the denominator can never equal 0. When you factor the numerator you get, (x-3)(x+1)/(x+1). You can cancel out the (x+1), so therefor you are left with (x-3). #Becasue you were able to cancel somthing out these means that there is a hole in the graph, which means there is a removable dicontinuity at x=-1. In order to find where the removable discontinuty is you should plug x=-1 into the new funcation. g(x) = (x-3) N(g(-1)) #This means the point of removeable diconsinuity is at (-1,-4) #What is also interesting to point out is that sage graphs does not show holes in the graph. plot((x^2 - 2*x -3)/(x+1), -2,0)
-3.90000000000000 -3.99000000000000 -3.99899999999997 -4.10000000000000 -4.01000000000000 -4.00099999999992 -4.00000000000000
#Question 5. In the iconic and legendary movie, Mean Girls, the main character Cady Heron, played by Lindsay Lohan, is on the Mathletes. During a Mathelete competition Cady is tasked with finding the limit as x approaches 0, of f(x) = (ln(1-x)-sinh(x))/(1-cosh(x)^2). She comes to the conclusion that "The Limit Does Not Exist". Please explain numerically, and then graphicly, why the limit does not exist. f(x) = (ln(1-x)-sinh(x))/(1-cosh(x)^2) N(f(-0.1)) N(f(-0.01)) N(f(-0.001)) #as x approaches 0 from the left side, f(x) approaches negative infinity N(f(0.1)) N(f(0.01)) N(f(0.001)) #as x approaches 0 from the right side, f(x) approaches positive infinity #Because the limits of f(x) as x approaches 0 from the left and right side are negative infinity and positive infinity respectively you can say that the limit of f(x), as x approaches 0, does not exist. This is also an example of infinite disontinuity. plot((ln(1-x)-sinh(x))/(1-cosh(x)^2), -0.01,0.01) #When you plot the graph you can see as x apporoaches 0 from the left side, f(x) will never touch x=0 and contunitue to negative infinity. Also when x apporoaches 0 from the right side, f(x) will never touch x=0 and continue to positive infinity. Becasue of this the limit of f(x) as x approaches 0, does not exist.
-19.4826641171976 -199.498325174270 -1999.49983261544 20.4843542802000 200.498341843273 2000.49983278190