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Project: Aurora Stefoff - Calculus 1 Lab Fall 2023

Path: 04 - Limits/Limits Assignment.ipynb

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Kernel: SageMath 10.0

Limits Assignment

Note:

You'll see comments below on Questions 1 through 3 stating, "The limit is approximately." Type a number at the end of this line to indicate your answer.

If the (two-sided) limit does not exist, write DNE.

Question 1

[2 points] Consider $\displaystyle\lim_{x\to-2}\frac{x^2-x-6}{x^3-6x^2+32}$ .

Caution: It's a little hard to see, but this limit is at $-2$ .

Part a

Estimate the limit to two decimal places by zooming in on a graph.

In [4]:

(-2) == (-2)^3 - 6(-2)^2 + 32(-2)^2 - (-2) - 6

Out[4]:

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
Cell In [4], line 1
----> 1 (-Integer(2)) == (-Integer(2))**Integer(3) - Integer(6)(-Integer(2))**Integer(2) + Integer(32)(-Integer(2))**Integer(2) - (-Integer(2)) - Integer(6)
TypeError: 'sage.rings.integer.Integer' object is not callable

In [0]:

#The limit is approximately:92

Part b

Estimate the limit numerically from the left using at least seven values.

In [0]:

In [0]:

#The limit is approximately:

Part c

Estimate the limit numerically from the right using at least seven values.

In [0]:

In [0]:

#The limit is approximately:

Part d

Compute the limit using CoCalc's limit command. [Convert your answer to a decimal in order to compare it with the results above.]

In [0]:

Question 2

[2 points] Consider $\displaystyle\lim_{x\to1}\frac{x^3-1}{\sqrt{x}-1}$ .

Part a

Estimate the limit to two decimal places by zooming in on a graph.

In [0]:

In [0]:

#The limit is approximately:

Part b

Estimate the limit numerically from the left using at least seven values.

In [0]:

In [0]:

#The limit is approximately:

Part c

Estimate the limit numerically from the right using at least seven values.

In [0]:

In [0]:

#The limit is approximately:

Part d

Compute the limit using CoCalc's limit command.

In [0]:

Question 3

[2 points] Consider $\displaystyle\lim_{x\to0}\frac{x\cdot (x+2)}{|x|}$ .

Note: To get $|x|$ , use abs(x) in CoCalc.

Part a

Estimate the limit to two decimal places by zooming in on a graph.

In [2]:

plot((x* (x+2) ) /abs (x) , xmin=-.1, xmax=.1)

Out[2]:

In [0]:

#The limit is approximately:DNE

Part b

Estimate the limit numerically from the left using at least seven values.

In [4]:

var('x')
f(x) = (x* (x+2) )/abs(x)
f(-.1)

Out[4]:

-1.90000000000000

In [7]:

var('x')
f (x) = (x* (x+2) )/abs (x)
f(- .1)

Out[7]:

-1.90000000000000

In [0]:

#The limit is approximately: -1.99

In [0]:

Part c

Estimate the limit numerically from the right using at least seven values.

In [8]:

var('x')
f(x) = (x* (x+2) ) /abs (x)
f(.1)

Out[8]:

2.10000000000000

In [9]:

var('x')
f(x) = (x* (x+2) ) /abs (x)
f(.01)

Out[9]:

2.01000000000000

In [0]:

#The limit is approximately: 2.00

Part d

Compute the limit using CoCalc's limit command.

In [10]:

var ('x')
f(x)=(x* (x+2)) / (abs (x))
limit(f(x),x=0)

Out[10]:

und

Question 4

[1 point] Let $\displaystyle f(x)=\frac{x^2-4}{x^2-1}$ .

Part a

Use CoCalc's limit command to compute the right limit $\displaystyle\lim_{x\to1^+}f(x)$ .

In [11]:

var('x')
f(x) = (x^2-4) /(x^2-1)
limit (f(x),x=1,dir="right")

Out[11]:

-Infinity

Part b

Use CoCalc's limit command to compute the left limit $\displaystyle\lim_{x\to1^-}f(x)$ .

[Careful: This is not the limit as x approaches $-1$ . This is the limit as x approaches positive $1$ from the left.]

In [12]:

var('x')
f(x) = (x^2-4) / (x^2-1)
limit(f(x),x=1,dir= 'left')

Out[12]:

+Infinity

Question 5

[2 points] Consider the function $\displaystyle f(x)=\frac{\sqrt{4x^2-9x}}{4x-1}$ .

Part a

Compute $\displaystyle\lim_{x\to-\infty}f(x)$ .

In [13]:

var('x')
f(x)=sqrt(4*x^2-9*x)/(4*x-1)
limit(f(x),x=-infinity)

Out[13]:

-1/2

Part b

Compute $\displaystyle\lim_{x\to\infty}f(x)$ .

In [15]:

var('x')
f(x) =sqrt (4*x^ 2-9*x) /(4*x-1)
limit(f(x),x=+infinity)

Out[15]:

1/2

Part c

Graph $f(x)$ using xmin= -100, xmax=100, ymin= -1, ymax=1.

Recall, if either of the limits above is a number, then the graph of $f$ has a horizontal asymptote. Does your graph have two horizontal asymptotes that match the answers from parts a and b?

In [17]:

plot(sqrt(4*x^2-9*x)/(4*x-1), xmin=-100,xmax=100, ymin=-1, ymax=1)

Out[17]:

Question 6

[1 point] Let $f(x)=\sqrt{2x+4}$ . Compute $\displaystyle\lim_{b\to a}\frac{f(b)-f(a)}{b-a}$ .

[Don't forget to declare variables a and b.]

In [18]:

var('x,a,b')
f(x)=sqrt(2*x+4)
limit ((f(b)-f (a))/(b-a), b=a)

Out[18]:

1/sqrt(2*a + 4)

In [19]:

Out[19]:

1

In [0]:

In [0]:

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Limits Assignment

Note:

Question 1

Part a

Part b

Part c

Part d

Question 2

Part a

Part b

Part c

Part d

Question 3

Part a

Part b

Part c

Part d

Question 4

Part a

Part b

Question 5

Part a

Part b

Part c

Question 6

Product

Resources

Company

Real-time collaboration for Jupyter Notebooks, Linux Terminals, LaTeX, VS Code, R IDE, and more, all in one place. Commercial Alternative to JupyterHub.

Limits Assignment

Note:

Question 1

Part a

Part b

Part c

Part d

Question 2

Part a

Part b

Part c

Part d

Question 3

Part a

Part b

Part c

Part d

Question 4

Part a

Part b

Question 5

Part a

Part b

Part c

Question 6

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