Elise T.
2/15/2016 01:27:32 pm
L'Hopital's Rule:
Paige Eber
2/16/2016 05:29:25 pm
Problem 18 from 8.1 is an example of this.
Abigail Chase
2/15/2016 02:00:04 pm
If the limit as X approaches infinity of f(x) over g(x) is a number that is not equal to zero, f(x) and g(x) grow at the same rate. (8.2)
Shreyas Mohan
2/15/2016 03:12:14 pm
Sometimes when you apply L'Hopital's rule to a rational function, you get the indeterminate form again. If this happens you can keep applying L'Hopital's rule until you get an expression where you can plug in the value the limit is approaching.
Allison Y
2/15/2016 03:40:53 pm
8.1 homework number 33:
Michelle H
2/15/2016 05:54:43 pm
You use L'Hopital's Rule when nothing can be divided out, substitution cannot be used, or when there is no graph provided
Shreyas Mohan
2/15/2016 06:00:54 pm
Shreyas Mohan
2/15/2016 06:02:52 pm
Sorry I accidentally hit submit without typing anything. Here is a list of some common indeterminate forms:
Jeffrey Willhauck
2/15/2016 06:10:55 pm
Here's a helpful picture for when you don't have a nice fraction for l'hôpitals. The first quiz problem is limited x> infinity xtan (1/x), which was a homework question (#24, 8.1)
N. Johnson
2/15/2016 06:31:19 pm
While I can appreciate your resourcefulness, we have quite a few exercises in our textbook. Can you choose one, work it out fully, and submit it to show that you know how to work through the exercises from the homework?
Jeffrey W
2/15/2016 10:36:07 pm
Here is an example 8 from the book. Let me know if you can't read anything.
Jeffrey W
2/15/2016 10:36:54 pm
Sorry, example 8 from section 8.1
Nathan Rao
2/15/2016 06:16:15 pm
Chapter 8.2 exercise #6
Nathan Rao
2/15/2016 06:19:33 pm
https://docs.google.com/document/d/1uVWveJmIV7X5IRELoZdZlBUNqlkJkxUfv6NmmT4iA70/pub
N. Johnson
2/15/2016 06:35:06 pm
Nice exercise. How do you know that x^2 grows faster? That should be included in your exercises because it will help when you are studying for the AP exam.
Nathan Rao
2/15/2016 10:43:23 pm
The reason we know that x^2 grows faster is because when the limit goes to zero that means that the function in the denominator grows faster. Since x^2 was the denominator that means that it increases faster than the function in the numerator and if the denominator grows at a faster rate than the numerator the difference between the two would keep growing and keep pushing the limit closer and closer to zero.
Kevin Knox
2/15/2016 06:24:57 pm
If a limit is lead to the indeterminate form of 1^infinity, 0^0, or infinity^0, it can sometimes be solved by taking the natural logarithm of the function and raising the constant 'e' to what was previously found.
N. Johnson
2/15/2016 06:37:05 pm
Can you include a specific example to further demonstrate that you understand what you wrote in theory? Thanks.
Kevin Knox
2/16/2016 05:37:13 pm
Sure thing!
Tanmayi K
2/15/2016 06:51:15 pm
As Shreyas said earlier, sometimes you have to apply L'Hopital's rule multiple times to solve for the limit. An example of this is number 6 in the 8.1 homework.
N. Johnson
2/15/2016 07:33:34 pm
Tanmayi,
Tanmayi K
2/15/2016 08:17:34 pm
Sorry about that, I think I made it accessible to anybody in ipsd instead of anyone with the link but I believe I fixed it
Rachel Willy
2/15/2016 06:51:53 pm
F grows at a faster rate than g when:
Taylor Garcia
2/15/2016 06:53:57 pm
If the limit as x approaches infinity of g(x) over f(x) is equal to 0 or if f(x) over g(x) approaches infinity, then that means the function f(x) grows faster than g(x)
Taylor Garcia
2/15/2016 06:55:23 pm
whoops, just ignore this
Taylor Garcia
2/15/2016 06:59:24 pm
When two functions grow at the same rate, they have a transitive relation. If f(x) grows at the same rate as g(x) and g(x) grows at the same rate as h(x), then f(x) grows at the same rate as h(x)
Christopher Glenn
2/15/2016 07:10:49 pm
I worked through # 27 from 8.1
N. Johnson
2/15/2016 07:45:33 pm
Are you sure that your derivative (step 2 in your example) is correct? Please place your work on the website for all to see, rather than going through google docs.
Brandon C
2/15/2016 08:06:25 pm
In your step 2, the derivative of ln(x) was correct but the derivative of ln(sinx) is not 1/sinx. you also have to multiply it by the derivative of the inside of the parentheses (Chain Rule). So the derivative of ln(sinx) is going to be (1/sinx)(cosx).
Christopher Glenn
2/15/2016 08:45:56 pm
Oh thanks Brandon. I knew I'd forget chain rule at some point. So that first derivative is (1/x)  (1/sinx)(cosx)). Then you make common denominators which turns it into (sinx  xcosx)/ xsinx. The limit as this approaches zero is still 0/0, so we have to take the derivative again. That gets us (xsinx)/(xcosx + sinx). This too is 0/0, so we derive one more time to get: (xcosx + sinx)/(xsinx +2cosx). The limit of this as x approaches 0 is 0/2 which is just 0.
Juan Guevara
2/15/2016 11:55:29 pm
(1) Use the logarithm quotient rule to write the problem in a single logarithm to give you ln (x/sinx) as the limit of x approaches 0+.
Kavya Anjur
2/15/2016 09:26:19 pm
This is problem #35 from 8.1
Ananth Putcha
2/15/2016 10:08:15 pm
When the limit as x approaches infinity of f(x)/g(x)=0, f is of smaller order than g. The notation for this is f=o(g).
Ananth Putcha
2/15/2016 10:15:26 pm
If f(x)/g(x) is less than or equal to a positive integer, M, then f is of "at most the order" of g as x approaches infinity for a sufficiently large value of x. The notation for this is f=O(g).
Shawn Park
2/15/2016 10:33:27 pm
I did Problem #27 from 8.1. Sorry for the long link.
N. Johnson
2/18/2016 11:45:21 am
I cannot access this response. Can you please repost it?
Juan Guevara
2/15/2016 10:46:41 pm
Section 8.1 L'Hôpital's Rule (p.424 #20)
N. Johnson
2/18/2016 11:46:47 am
I cannot access this site. Please repost it with google docs and open the access up to EVERYONE.
Juan Guevara
2/15/2016 10:52:36 pm
First link did not work, so here's the new one:
Rumi V
2/15/2016 11:19:11 pm
8.1 p. 424 #35
Shawn Park
2/15/2016 11:58:24 pm
I also did #31 from 8.1
N. Johnson
2/18/2016 11:49:24 am
I cannot access this. Please repost it in google doc (not mail) so that everyone can see it.
Shawn Park
2/18/2016 10:28:00 pm
I'll just type this out!
Allison Y
2/16/2016 11:25:18 am
8.2 #15
Elise T
2/16/2016 02:31:15 pm
Here's 8.2 #21!
Rachel Willy
2/16/2016 03:00:50 pm
https://docs.google.com/document/d/1TAY2sQB4EMeal2F9EhZDoRb1AxnmCVQDNBp6n6UB0ms/edit?usp=sharing
Nathan Rao
2/16/2016 03:22:37 pm
8.2 Exercise # 23
Christopher Glenn
2/16/2016 04:11:40 pm
I'll do 8.2 #9
Shreyas Mohan
2/16/2016 04:44:32 pm
Alright so I have a question. For 8.2 #3, it asks you to determine whether 4^x grows faster/slower/same as e^x. I can reason through it and know that 4^x grows faster since 4>e, but is there a way to algebraically prove it? I tried applying l'hopital but I didn't really get anywhere.
Abigail Chase
2/16/2016 05:08:13 pm
Yes, using L'Hopital's the limit as x approaches infinity of 4^x over e^x is equal to the limit as x approaches infinity of (4/e)^x which is infinity and since 4/e > 1, 4^x grows faster than e^x. Another way you could determine this is because 4>e because e= about 2.718.
N. Johnson
2/18/2016 11:36:16 am
Nice job, Abigail!
Kevin Knox
2/16/2016 05:54:34 pm
Transitivity of growing rates:
Juan Guevara
2/16/2016 11:34:50 pm
“Growing at the same rate” is transitive.
Rumi Venkatesh
2/16/2016 07:19:19 pm
https://drive.google.com/a/k12.ipsd.org/file/d/0B5ysYKgGu_bsakZvVkl3Uno4UGM/view?usp=docslist_api
Taylor Garcia
2/16/2016 08:06:10 pm
8.2 #7
Michelle H
2/16/2016 08:09:37 pm
One can only use L'Hospital's rule when it is in fraction format. When you are stuck on a problem, try to make the limit into one function with a specific f(x) in the numerator and a specific g(x) in the denominator.
Ananth Putcha
2/16/2016 09:38:12 pm
A simple example of this would be if f(x)=2(x^2) + x then you can factor out the x so you have x(2x+1) which can be changed into the fraction (2x+1)/(1/x). Now you can use f(x)=2x+1 and g(x)=1/x to use l'hopital's rule.
N.Johnson
2/18/2016 11:37:47 am
Nice addon, Ananth!
Kavya Anjur
2/16/2016 08:30:59 pm
In section 8.2, using the bigoh notation helps with binary and sequential searches. A sequential search takes O(n) steps while a binary search takes O(log base 2 of n) steps. The difference between the two types of searches is that the sequential method does not make use of the alphabetical arrangement of words while the binary search method does, and efficiently eliminates about half of the words in the process. If you were looking at a list with 26,000 words, it might take 26,000 steps with sequential but only 15 with binary [as 15= O(log base 2 of 26,000)].
N. Johnson
2/18/2016 11:39:16 am
What? How does this relate to the homework? Can you provide an example from our text to further explain what you mean? Thanks.
Tanmayi K
2/16/2016 09:21:20 pm
By applying the definitions of faster, slower, and same rate growth as x approaches infinity as well as the definition of f of smaller order than g, one can solve problem 32 by analyzing the graph. The graphs is of f/g. The graph of f/g approaches 0 as x approaches infinity. Now, using the definition of the o notation, we know that the limit of f/g as x approaches infinity can be written as f=o(g) or f is littleoh of fog so we can apply this to problem 32. Therefore only part 'i' is true.
Tanmayi K
2/16/2016 09:22:37 pm
*The graph not The graphs
N. Johnson
2/18/2016 11:40:40 am
What is part "I"?
Jeffrey W
2/16/2016 09:43:36 pm
Here's the answer to 8.1 #57
Jeffrey W
2/16/2016 09:44:23 pm
Sorry, it's number #52, not #57 2/16/2016 09:58:46 pm
8.2 p.431 #28
Shawn Park
2/16/2016 10:04:12 pm
Hi guys...
Shawn Park
2/16/2016 11:38:43 pm
Try this link
Herven Barham
2/16/2016 10:14:11 pm
Chapter 8.1 Problem #39
Paige Eber
2/16/2016 10:14:51 pm
Heres #22 from 8.2
Juan Guevara
2/16/2016 11:05:31 pm
Section 8.2 Relative Rates of Growth (p. 331 #27)
Juan Guevara
2/16/2016 11:24:55 pm
Definition 1: Improper Integrals with Infinite Integration Limits
Juan Guevara
2/16/2016 11:28:13 pm
These definitions are from Section 8.3 Improper Integrals.
Nathan Rao
2/17/2016 08:47:46 am
8.2 Exercise # 31
Elise T
2/17/2016 10:21:57 am
Here are the basic limits that we reviewed today to help with this chapter! (there's a link to a picture written out at the bottom if that's easier for anyone)
Abigail Chase
2/17/2016 02:18:42 pm
L'Hopitals was actually created by John Bernoulli, but L'Hopital took the credit. L'Hopital made the first calculus textbook.
Michelle H
2/17/2016 03:16:42 pm
When dealing with exponents, you take the natural log of both sides. Since you did this, you must also take the limit as x> a of both the left and right side as well.
Rachel Willy
2/17/2016 04:24:03 pm
Any constant in the numerator divided by 0 in the denominator goes to ∞. Here's some examples with the graph of 1/x!
Shreyas Mohan
2/17/2016 04:59:26 pm
Took a look at the beginning of section 8.3 and saw some pretty interesting stuff. It talks about integrals where the bounds go to infinity. We might think that this means the area under the curve is infinite, but in some cases it can be finite.
Ananth Putcha
2/17/2016 05:05:04 pm
Here are some basic limits that we should all memorize...
Ananth Putcha
2/17/2016 05:06:00 pm
Oh sorry, I didn't see that Elise already posted these
Allison Y
2/17/2016 05:18:06 pm
8.2 #16
Tanmayi K
2/17/2016 06:07:39 pm
8.2 #18
Paige Eber
2/17/2016 06:07:58 pm
Another L'hopital practice problem; 8.1 #5
Brandon C
2/17/2016 08:05:54 pm
8.1 #42
Elise T
2/17/2016 08:51:30 pm
8.2 #13
Christopher Glenn
2/17/2016 09:11:06 pm
I'll do a problem from the future 8.3. 2/17/2016 09:24:08 pm
8.2 p.432 #42
Jeffrey W
2/17/2016 09:24:39 pm
So can we just assume like in question number 9 of the 8.2 homework, a constant (C>1) raised to the power of X will always grow faster than X raised to a power, no matter how ridiculously high the power may be?
Shawn Park
2/17/2016 10:11:21 pm
EVENTUALLY, Yes. When looking at end behavior, the constant raised to a power will grow faster than x raised to a power.
Shawn Park
2/17/2016 10:07:33 pm
#26 from 8.2
Rumi Venkatesh
2/17/2016 10:13:59 pm
Comparing x^33x+1 to e^x
Juan Guevara
2/17/2016 10:20:15 pm
Section 8.2 Relative Rates of Growth (p. 333 #43)
Tanmayi K
2/17/2016 10:21:20 pm
Hey guys so I'm not sure if anyone has posted about #30 from 8.2 yet but I have a question about part d.
Abigail Chase
2/18/2016 01:51:09 pm
In this case M is 3/2. Part d) is true because the limit of (2+cosx)/2 as x approaches infinity is 3/2. Therefore 2+cosx=O(2) is true because lim x> infinity of (2+cosx)/2 is equal to M (which is 3/2).
Ananth Putcha
2/17/2016 10:24:55 pm
8.2 #12: Determine whether x^3 + 3 grows faster, at the same rate, or slower than e^x as x>∞.
Herven
2/17/2016 10:32:11 pm
8.2 #17
Herven
2/18/2016 08:18:40 pm
I typed out my proof so its easier to see
Kevin Knox
2/17/2016 11:22:20 pm
8.2 #8
Kavya Anjur
2/17/2016 11:48:09 pm
For 8.2, problem #25, one can see how the Transitivity of growing rates can be applied. First, start by comparing f1 to f2. One would take the limit as x approaches infinity of f2/f1, which is equal to [ the square root of (10x+1)] / [the square root of x]. Essentially this can be simplified to the square root of [10 + (1/x)], and when one takes the limit of that going to infinity, then they get the square root of 10, which is a constant, thus showing that f1 and f2 grow at the same rate. The next step is to compare f3 and f1 by taking the limit as x approaches infinity of f3/f1, which is [the square root of (x+1)] / [the square root of x]. This simplifies to the square root of [1+ (1/x)]. The limit of this function as x approaches infinity is 1, showing that f1 and f3 grow at the same rate. Thus, by transitivity, f2 and f3 grow at the same rate and so does f1 and f2, so all three functions grow at the same rate as x approaches infinity.
Kavya Anjur
2/18/2016 12:07:41 am
For 8.1, problem #22, one must take the limit as y approaches (pi/2) of the function [(pi/2)  y][tan(y)]. One can rewrite tan(y) as sin(y)/cos(y), thus taking limit as y approaches (pi/2) of the function [((pi/2)  y)(sin(y))] / [cos(y)]. Using l'hopitals rule, one must take the derivative of the numerator over the denominator, resulting in limit as y approaches (pi/2) of [((pi/2)  y)(cos(y)) + (sin(y))(1)] / [sin(y)]. Taking this limit gives one the solution of [((pi/2)(pi/2))(cos(pi/2)) + (sin(pi/2))(1)] / [sin(pi/2)], which equals [(1)(1)] / [1]. Therefore, the solution is 1.
Nathan Rao
2/18/2016 08:34:34 am
Indeterminate Form [infinty]^0
Elise T
2/18/2016 10:32:26 am
In order to prove your work when working with the indeterminate form 0^0, this second step needs to be written out:
Abigail Chase
2/18/2016 02:04:00 pm
I have a question about 8.2 #30, part g). It asks to show whether ln(lnx)=O(lnx) so first I set up my fraction which was (ln(lnx))/lnx then I took the derivative (L'Hopitals rule) and I got ((1/lnx)(1/x))/(1/x) which simplifies to the limit as x> infinity of 1/lnx which is zero. The answer says that it is true because 0 is less than or equal to 1. However I thought that the answer should be little "o" because I got zero as my answer and not a number "M". Can someone explain why it is "O" (big o)?
Rachel Willy
2/18/2016 03:18:06 pm
In class today we learned how to find the limit of the indeterminate form of 0^0 so here is an example using that form!
Paige Eber
2/18/2016 03:30:23 pm
8.1 and 8.2 connect through limits. In 8.1 we learned to take difficult limits with L'Hopital's Rule, which states that the limit as x approaches a of f(x)/g(x) = f'(a)/g'(a)
Allison Y
2/18/2016 05:49:55 pm
8.1 #38
Jeffrey W
2/18/2016 09:34:16 pm
Here's a little sneak peak of 8.3
Jeffrey W
2/18/2016 09:38:08 pm
Sorry, for 2. the example should be the lim b> infinity, not lim x> infinity
.
2/18/2016 10:13:01 pm
Lim x (1 + 2x) 1/ (2lnx)
Juan Guevara
2/18/2016 10:43:29 pm
_______________________________________________________
Shawn Park
2/18/2016 10:36:43 pm
#23 of 8.1
Nathan Rao
2/19/2016 09:04:12 am
8.3 Exercise # 11
Elise T
2/19/2016 10:31:57 am
There are 3 steps to help determine convergence or divergence of an improper integral:
Michelle H
2/19/2016 11:38:42 am
It is an improper integral when the bounds include infinity or negative infinity. In this case, substitute the infinity bound with "a" or "b" to make it a definite integral.
Abigail Chase
2/19/2016 01:22:16 pm
Example Problem. Determine if the integral from0 to infinity of 1/e^x dx converges or diverges. Convert this equation by making "b" infinity. The limit as b goes to infinity of the integral of 0 to b of 1/e^x is solved by taking the antiderivative of 1/e^x which is 1/e^x and evaluating the limit as b goes to infinity of (1/e^b  (1/e^0)) and because the limit as b goes to infinity of 1/e^b is 1, the function converges.
Kavya Anjur
2/19/2016 02:44:53 pm
If an improper integral converges, then that means that it comes together visually on the graph. Mathematically, the limit of the improper integral will be finite if it converges, and the limit will be the value of the improper integral. If it diverges, then the limit of the improper integral does not exist. Regarding divergent integrals, the limit is not numerically possible, thus it fails to exist.
Kavya Anjur
2/19/2016 02:50:15 pm
These are only regarding improper integrals with infinite discontinuities. An improper integral is an integral of a function that becomes infinite at a point within the interval of integration. 2/19/2016 06:01:27 pm
p.443 8.3 #52
Shawn Park
2/19/2016 07:47:28 pm
#9 from 8.3 P. 442
Kevin Knox
2/19/2016 09:15:09 pm
8.2 #5
Allison Y
2/19/2016 09:51:56 pm
8.3 #7
Allison Y
2/20/2016 04:10:50 pm
8.2 #20
Nathan Rao
2/22/2016 09:08:13 am
Section 8.3 Exercise #35
Abigail Chase
2/22/2016 11:23:49 am
For partial fraction decomposition, you start by decomposing the denominator and factoring the denominator.
Elise T
2/22/2016 03:59:15 pm
8.3 #6 (link to picture of work: http://tinyurl.com/jzpvnkd)
Ananth Putcha
2/22/2016 10:58:03 pm
8.3 #3: Integral from 8 to 1 of dx/(x^1/3)
Kevin Knox
2/22/2016 11:10:54 pm
8.3 #2
Elise T
2/23/2016 10:34:34 am
Limit Comparison Test:
Abigail Chase
2/23/2016 11:09:57 am
Remember that the limit as x approaches infinity for both inverse secant and inverse tangent is pi/2. 2/23/2016 09:50:55 pm
8.4 p.452 #8
Shawn Park
2/23/2016 09:52:08 pm
#18 from 8.3
Ananth Putcha
2/23/2016 10:07:02 pm
8.3 #12: integral from infinity to 2 of (2dx/(x^2+4))
Christopher Glenn
2/23/2016 11:17:30 pm
I chose to do #3 from 8.4. The initial equation is (t+1)/(t1)(t^2) and you have to make a partial fraction out of it. so you factor it but because of (t^2) you have to make two parts out of that: t and t^2. From that you get A/t + B/t^2 + C/(t1) =(t+1)/(t1)(t^2)
Kevin Knox
2/23/2016 11:56:47 pm
8.3 #1
Nathan Rao
2/24/2016 08:57:09 am
Section 8.4 #27
Jeffrey W
2/24/2016 06:31:46 pm
When solving an improper integral, it's important to note which bound you are replacing so that you can look at the limit at the point from the correct side. For instance, if the integral is
Elise T
2/24/2016 07:51:35 pm
8.1 #17 (picture: http://tinyurl.com/hpqoagv)
Michelle H
2/24/2016 08:00:23 pm
When approached with a limit, you have a top and bottom bound. When it is a specific number "a", you must specify if it is approached from the left or right. 2/24/2016 10:22:50 pm
Shortcut for evaluating the integral from zero to infinity of 1/x^p where p is a constant:
Brandon C
2/24/2016 10:43:01 pm
An example of this would be if:
Shawn Park
2/24/2016 10:36:17 pm
Chapter 8 Review #13
Christopher Glenn
2/24/2016 10:47:22 pm
Chapter 8 Review #10
Kavya Anjur
2/24/2016 11:06:39 pm
In 8.3 #5, there is an infinite discontinuity when x=0, which is why it involves improper integrals. You start by taking the limit from the right of b to 0 of the integral of ( x^2)(e^[1/x]) from ln2 to b. You can solve the integral by doing u substitution and getting the integral from 1/b to 1/ln2 of e^u. Once you take the limit of this integral as b approaches 0 (solving the integral with an antiderivative), you get e^(1/ln2) + e^infinity which is equal to infinity. Thus, the integral converges.
Herven
2/24/2016 11:41:32 pm
Chapter 8 Review #3
Kevin Knox
2/24/2016 11:52:40 pm
8.4 #7
Kavya Anjur
2/25/2016 08:52:35 pm
In 8.4 #6, one breaks it up into the functions of y^3 +1,and y^2 +4. It is then necessary to do long division to further simplify. When you divide y^3 +1 by y^2 +4 you get y as a result with a remainder of (4y+1)> which has the denominator of y^2 +4 because it is a remainder. Thus, the function of (y^3 +1) / (y^2 +4) = y + (4y+1)/(y^2 +4). Since it is possible to decompose this function any further, this is your final answer.
Kevin Knox
2/25/2016 09:35:23 pm
8.1 #15
Kevin Knox
2/26/2016 10:10:15 pm
8.2 #20
Shawn Park
3/1/2016 08:49:13 pm
BC Calculus 8.18.2 Short Quiz #1
Allison Y
3/9/2016 06:57:18 pm
8.2 #4
Shawn Park
3/9/2016 08:19:16 pm
Semester 1 Calculus Review #18
Allison Y
3/11/2016 03:34:04 pm
8.1 #26 Comments are closed.

AuthorMrs. Johnson's 20152016 BC Calculus Center for Review. By participating in this blog, you are indicating that the work that you submit is your own. If found to be otherwise true, you will not receive credit. Happy blogging!
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