- Re: [gsu2211] Sunday by Sheldon Elias Schiffer
- Re: [gsu2211] Sunday by richard hollis
- Re: [gsu2211] Sunday by Sheldon Elias Schiffer
- Re: [gsu2211] Sunday by Sheldon Elias Schiffer
- Re: [gsu2211] Sunday by Seth Archer
- Re: [gsu2211] Sunday by Sheldon Elias Schiffer
- Re: [gsu2211] Sunday by Seth Archer

- Re: [gsu2211] Sunday by Sheldon Elias Schiffer
- Re: [gsu2211] Sunday by Lauren Peyton
- Re: [gsu2211] Sunday by Sheldon Elias Schiffer
- Re: [gsu2211] Sunday by Seth Archer

- Re: [gsu2211] Sunday by Seth Archer
- Re: [gsu2211] Sunday by Seth Archer

- From:
- Sheldon Elias Schiffer
- Date:
- 2013-02-10 @ 16:52

I'm free still for the library meeting, if anyone is up for it. If we go for a virtual, I can be on Skype at 6:30 PM. My skype name is: sheldonelias I think it could work well on Skype. I have done so programming jam session on Skype. Otherwise, I had some questions about two exercises on P.63. #49 & #57. I didn't see the factoring pattern. I know what to do once I can get rid of the radicals, but getting there was a little mysterious. The conjugate alone is not the way, but probably is one of the steps. Here are the questions: 49. lim x>1 of x^2 + 1/(sqrt 2x+2) - 2 57. lim x>2 of [(sqrt 6-x) - x ] / x - 2 If anyone has the factoring trick up their sleeve, remind me. It's not coming back so far. Thanks, -S ________________________________ From: gsu2211@librelist.com [gsu2211@librelist.com] on behalf of Seth Archer [learc83@gmail.com] Sent: Sunday, February 10, 2013 11:30 AM To: gsu2211@librelist.com Subject: [gsu2211] Sunday Since almost everyone is having a hard time making it tonight. Why don't we just have a virtual study group. Anyone who has any remaining questions about anything, just post them to the mailing list. -Seth

- From:
- richard hollis
- Date:
- 2013-02-11 @ 03:07

Hey going to be at math lab most of tomorrow if anybody wants to review Sent from my Windows Phone ________________________________ From: Sheldon Elias Schiffer<mailto:schiffer@gsu.edu> Sent: 2/10/2013 9:38 PM To: gsu2211@librelist.com<mailto:gsu2211@librelist.com> Subject: RE: [gsu2211] Sunday I get the idea, but the answers don't make sense. Where are the new functions being defined to yield a 1 and a 6 to make 1/6 (#41) or a 1 and a 3 to make 1/3 (#43)? -S ________________________________ From: gsu2211@librelist.com [gsu2211@librelist.com] on behalf of Seth Archer [learc83@gmail.com] Sent: Sunday, February 10, 2013 9:00 PM To: gsu2211@librelist.com Subject: Re: [gsu2211] Sunday They took the limit at 5. Then if you set f(5) to be equal to the limit at 5, the function becomes continuous at f(5). However we haven't learned how to take limits from expressions like that. The only thing you could do using what we've learned is to plug in numbers close to 5. On Sun, Feb 10, 2013 at 8:19 PM, Sheldon Elias Schiffer <schiffer@gsu.edu<mailto:schiffer@gsu.edu>> wrote: Seth, or anyone else, do you have a strategy for P. 90 #41 & 43? I get the idea of composite limits, but the question is, make the function continuous at 5. The answer is 1/6 and 1/3? 41. f(x)= [(sqrt (x+4)) - 3] / (x - 5) 43. f(x) = [sqrt(2x - 1) - 3] / (x - 5) Any idea how the book got that answer? -Sheldon

- From:
- Sheldon Elias Schiffer
- Date:
- 2013-02-10 @ 20:27

```
Ok. How about P. 72 #33.
I started to try it, but I don't know how solve f(c)=L+epsilon where f(c)
for (sin3c)/c = 3.25
The first of the two given epsilons is .25
My first move was to do the usual:
Solve for c. My guess is that since sin oscillates, there is not one
solution, but a solution within each rotation of the unit circle, so you
can set the equation so that sin3c = 0 or 1, and solve for c. Am I on to
something?
29 and 31 were easy. But bring in the trig functions with the pure algebra
we did not do in class.
-Sheldon
________________________________
From: gsu2211@librelist.com [gsu2211@librelist.com] on behalf of Seth
Archer [learc83@gmail.com]
Sent: Sunday, February 10, 2013 12:55 PM
To: gsu2211@librelist.com
Subject: Re: [gsu2211] Sunday
I'm about 99% sure there is no way to do 57 algebraically. Looks like you
have to use more advanced methods we haven't learned yet, I'm assuming
that's why the book suggested a graphing utility.
On Sun, Feb 10, 2013 at 12:34 PM, Seth Archer
<learc83@gmail.com<mailto:learc83@gmail.com>> wrote:
>49. lim x>1 of x^2 + 1/(sqrt 2x+2) - 2
The answer in the back of the book is wrong. The limit is infinity, so
DNE. If you plug in 1, you get 2/0. I also double checked it with my nice
calculator and it gave me infinity as well.
Still working on 57. The book answer is correct on that one, though.
On Sun, Feb 10, 2013 at 11:52 AM, Sheldon Elias Schiffer
<schiffer@gsu.edu<mailto:schiffer@gsu.edu>> wrote:
I'm free still for the library meeting, if anyone is up for it.
If we go for a virtual, I can be on Skype at 6:30 PM. My skype name is:
sheldonelias
I think it could work well on Skype. I have done so programming jam
session on Skype.
Otherwise, I had some questions about two exercises on P.63. #49 & #57. I
didn't see the factoring pattern. I know what to do once I can get rid of
the radicals, but getting there was a little mysterious. The conjugate
alone is not the way, but probably is one of the steps.
Here are the questions:
49. lim x>1 of x^2 + 1/(sqrt 2x+2) - 2
57. lim x>2 of [(sqrt 6-x) - x ] / x - 2
If anyone has the factoring trick up their sleeve, remind me. It's not
coming back so far.
Thanks,
-S
________________________________
From: gsu2211@librelist.com<mailto:gsu2211@librelist.com>
[gsu2211@librelist.com<mailto:gsu2211@librelist.com>] on behalf of Seth
Archer [learc83@gmail.com<mailto:learc83@gmail.com>]
Sent: Sunday, February 10, 2013 11:30 AM
To: gsu2211@librelist.com<mailto:gsu2211@librelist.com>
Subject: [gsu2211] Sunday
Since almost everyone is having a hard time making it tonight. Why don't
we just have a virtual study group.
Anyone who has any remaining questions about anything, just post them to
the mailing list.
-Seth
```

- From:
- Sheldon Elias Schiffer
- Date:
- 2013-02-11 @ 01:19

```
Seth, or anyone else, do you have a strategy for P. 90 #41 & 43?
I get the idea of composite limits, but the question is, make the function
continuous at 5. The answer is 1/6 and 1/3?
41. f(x)= [(sqrt (x+4)) - 3] / (x - 5)
43. f(x) = [sqrt(2x - 1) - 3] / (x - 5)
Any idea how the book got that answer?
-Sheldon
________________________________
From: gsu2211@librelist.com [gsu2211@librelist.com] on behalf of Lauren
Peyton [laurenpeyt@gmail.com]
Sent: Sunday, February 10, 2013 7:27 PM
To: gsu2211@librelist.com
Subject: Re: [gsu2211] Sunday
Hey guys! Sorry, I had to work late and the weather is terrible so I won't
be coming down.
I have a break tomorrow from 3-5:30 if anyone wants to meet up and review!
On Feb 10, 2013, at 4:26 PM, Sheldon Elias Schiffer
<schiffer@gsu.edu<mailto:schiffer@gsu.edu>> wrote:
So it's for real nobody can come to GSU to study? Let me know if someone
is coming. I can be there in 15 mins.
-S
________________________________
From: gsu2211@librelist.com<mailto:gsu2211@librelist.com>
[gsu2211@librelist.com<mailto:gsu2211@librelist.com>] on behalf of Sheldon
Elias Schiffer [schiffer@gsu.edu<mailto:schiffer@gsu.edu>]
Sent: Sunday, February 10, 2013 3:27 PM
To: gsu2211@librelist.com<mailto:gsu2211@librelist.com>
Subject: RE: [gsu2211] Sunday
Ok. How about P. 72 #33.
I started to try it, but I don't know how solve f(c)=L+epsilon where f(c)
for (sin3c)/c = 3.25
The first of the two given epsilons is .25
My first move was to do the usual:
Solve for c. My guess is that since sin oscillates, there is not one
solution, but a solution within each rotation of the unit circle, so you
can set the equation so that sin3c = 0 or 1, and solve for c. Am I on to
something?
29 and 31 were easy. But bring in the trig functions with the pure algebra
we did not do in class.
-Sheldon
________________________________
From: gsu2211@librelist.com<mailto:gsu2211@librelist.com>
[gsu2211@librelist.com<mailto:gsu2211@librelist.com>] on behalf of Seth
Archer [learc83@gmail.com<mailto:learc83@gmail.com>]
Sent: Sunday, February 10, 2013 12:55 PM
To: gsu2211@librelist.com<mailto:gsu2211@librelist.com>
Subject: Re: [gsu2211] Sunday
I'm about 99% sure there is no way to do 57 algebraically. Looks like you
have to use more advanced methods we haven't learned yet, I'm assuming
that's why the book suggested a graphing utility.
On Sun, Feb 10, 2013 at 12:34 PM, Seth
Archer<learc83@gmail.com<mailto:learc83@gmail.com>> wrote:
>49. lim x>1 of x^2 + 1/(sqrt 2x+2) - 2
The answer in the back of the book is wrong. The limit is infinity, so
DNE. If you plug in 1, you get 2/0. I also double checked it with my nice
calculator and it gave me infinity as well.
Still working on 57. The book answer is correct on that one, though.
On Sun, Feb 10, 2013 at 11:52 AM, Sheldon Elias
Schiffer<schiffer@gsu.edu<mailto:schiffer@gsu.edu>> wrote:
I'm free still for the library meeting, if anyone is up for it.
If we go for a virtual, I can be on Skype at 6:30 PM. My skype name is:
sheldonelias
I think it could work well on Skype. I have done so programming jam
session on Skype.
Otherwise, I had some questions about two exercises on P.63. #49 & #57. I
didn't see the factoring pattern. I know what to do once I can get rid of
th e radicals, but getting there was a little mysterious. The conjugate
alone is not the way, but probably is one of the steps.
Here are the questions:
49. lim x>1 of x^2 + 1/(sqrt 2x+2) - 2
57. lim x>2 of [(sqrt 6-x) - x ] / x - 2
If anyone has the factoring trick up their sleeve, remind me. It's not
coming back so far.
Thanks,
-S
________________________________
From:
gsu2211@librelist.com<mailto:gsu2211@librelist.com>[gsu2211@librelist.com<mailto:gsu2211@librelist.com>]
on behalf of Seth Archer [learc83@gmail.com<mailto:learc83@gmail.com>]
Sent: Sunday, February 10, 2013 11:30 AM
To: gsu2211@librelist.com<mailto:gsu2211@librelist.com>
Subject: [gsu2211] Sunday
Since almost everyone is having a hard time making it tonight. Why don't
we just have a virtual study group.
Anyone who has any remaining questions about anything, just post them to
the mailing list.
-Seth
```

- From:
- Seth Archer
- Date:
- 2013-02-11 @ 02:00

They took the limit at 5. Then if you set f(5) to be equal to the limit at 5, the function becomes continuous at f(5). However we haven't learned how to take limits from expressions like that. The only thing you could do using what we've learned is to plug in numbers close to 5. On Sun, Feb 10, 2013 at 8:19 PM, Sheldon Elias Schiffer <schiffer@gsu.edu>wrote: > Seth, or anyone else, do you have a strategy for P. 90 #41 & 43? > > I get the idea of composite limits, but the question is, make the > function continuous at 5. The answer is 1/6 and 1/3? > > 41. f(x)= [(sqrt (x+4)) - 3] / (x - 5) > > 43. f(x) = [sqrt(2x - 1) - 3] / (x - 5) > > Any idea how the book got that answer? > > -Sheldon >

- From:
- Sheldon Elias Schiffer
- Date:
- 2013-02-11 @ 03:32

1/6 isn't random though. Any idea how they got that? I am pretty sure this won't be on the test, so if you have better things to do, it can wait. -S ________________________________ From: gsu2211@librelist.com [gsu2211@librelist.com] on behalf of Seth Archer [learc83@gmail.com] Sent: Sunday, February 10, 2013 10:30 PM To: gsu2211@librelist.com Subject: Re: [gsu2211] Sunday They are just redefining the function into a piecewise function so the new definition is f(x) = [(sqrt (x+4)) - 3] / (x - 5), for x not equal to 5, and 1/6 for x equal to 5. On Sun, Feb 10, 2013 at 9:38 PM, Sheldon Elias Schiffer <schiffer@gsu.edu<mailto:schiffer@gsu.edu>> wrote: I get the idea, but the answers don't make sense. Where are the new functions being defined to yield a 1 and a 6 to make 1/6 (#41) or a 1 and a 3 to make 1/3 (#43)? -S ________________________________ From: gsu2211@librelist.com<mailto:gsu2211@librelist.com> [gsu2211@librelist.com<mailto:gsu2211@librelist.com>] on behalf of Seth Archer [learc83@gmail.com<mailto:learc83@gmail.com>] Sent: Sunday, February 10, 2013 9:00 PM To: gsu2211@librelist.com<mailto:gsu2211@librelist.com> Subject: Re: [gsu2211] Sunday They took the limit at 5. Then if you set f(5) to be equal to the limit at 5, the function becomes continuous at f(5). However we haven't learned how to take limits from expressions like that. The only thing you could do using what we've learned is to plug in numbers close to 5. On Sun, Feb 10, 2013 at 8:19 PM, Sheldon Elias Schiffer <schiffer@gsu.edu<mailto:schiffer@gsu.edu>> wrote: Seth, or anyone else, do you have a strategy for P. 90 #41 & 43? I get the idea of composite limits, but the question is, make the function continuous at 5. The answer is 1/6 and 1/3? 41. f(x)= [(sqrt (x+4)) - 3] / (x - 5) 43. f(x) = [sqrt(2x - 1) - 3] / (x - 5) Any idea how the book got that answer? -Sheldon

- From:
- Seth Archer
- Date:
- 2013-02-11 @ 05:10

>1/6 isn't random though. Any idea how they got that? Yeah, it's the limit of the function as x approaches 5. There's no way to get it algebraically. You have to use l'Hôpital's rule I believe. On Sun, Feb 10, 2013 at 10:32 PM, Sheldon Elias Schiffer <schiffer@gsu.edu>wrote: > 1/6 isn't random though. Any idea how they got that? > > I am pretty sure this won't be on the test, so if you have better things > to do, it can wait. > > -S > > ------------------------------ > *From:* gsu2211@librelist.com [gsu2211@librelist.com] on behalf of Seth > Archer [learc83@gmail.com] > *Sent:* Sunday, February 10, 2013 10:30 PM > > *To:* gsu2211@librelist.com > *Subject:* Re: [gsu2211] Sunday > > They are just redefining the function into a piecewise function so the > new definition is f(x) = [(sqrt (x+4)) - 3] / (x - 5), for x not equal > to 5, and 1/6 for x equal to 5. > > > On Sun, Feb 10, 2013 at 9:38 PM, Sheldon Elias Schiffer <schiffer@gsu.edu>wrote: > >> I get the idea, but the answers don't make sense. Where are the new >> functions being defined to yield a 1 and a 6 to make 1/6 (#41) or a 1 and a >> 3 to make 1/3 (#43)? >> >> -S >> ------------------------------ >> *From:* gsu2211@librelist.com [gsu2211@librelist.com] on behalf of Seth >> Archer [learc83@gmail.com] >> *Sent:* Sunday, February 10, 2013 9:00 PM >> >> *To:* gsu2211@librelist.com >> *Subject:* Re: [gsu2211] Sunday >> >> They took the limit at 5. Then if you set f(5) to be equal to the >> limit at 5, the function becomes continuous at f(5). However we haven't >> learned how to take limits from expressions like that. The only thing you >> could do using what we've learned is to plug in numbers close to 5. >> >> On Sun, Feb 10, 2013 at 8:19 PM, Sheldon Elias Schiffer <schiffer@gsu.edu >> > wrote: >> >>> Seth, or anyone else, do you have a strategy for P. 90 #41 & 43? >>> >>> I get the idea of composite limits, but the question is, make the >>> function continuous at 5. The answer is 1/6 and 1/3? >>> >>> 41. f(x)= [(sqrt (x+4)) - 3] / (x - 5) >>> >>> 43. f(x) = [sqrt(2x - 1) - 3] / (x - 5) >>> >>> Any idea how the book got that answer? >>> >>> -Sheldon >>> >> >> >

- From:
- Sheldon Elias Schiffer
- Date:
- 2013-02-10 @ 21:26

```
So it's for real nobody can come to GSU to study? Let me know if someone
is coming. I can be there in 15 mins.
-S
________________________________
From: gsu2211@librelist.com [gsu2211@librelist.com] on behalf of Sheldon
Elias Schiffer [schiffer@gsu.edu]
Sent: Sunday, February 10, 2013 3:27 PM
To: gsu2211@librelist.com
Subject: RE: [gsu2211] Sunday
Ok. How about P. 72 #33.
I started to try it, but I don't know how solve f(c)=L+epsilon where f(c)
for (sin3c)/c = 3.25
The first of the two given epsilons is .25
My first move was to do the usual:
Solve for c. My guess is that since sin oscillates, there is not one
solution, but a solution within each rotation of the unit circle, so you
can set the equation so that sin3c = 0 or 1, and solve for c. Am I on to
something?
29 and 31 were easy. But bring in the trig functions with the pure algebra
we did not do in class.
-Sheldon
________________________________
From: gsu2211@librelist.com [gsu2211@librelist.com] on behalf of Seth
Archer [learc83@gmail.com]
Sent: Sunday, February 10, 2013 12:55 PM
To: gsu2211@librelist.com
Subject: Re: [gsu2211] Sunday
I'm about 99% sure there is no way to do 57 algebraically. Looks like you
have to use more advanced methods we haven't learned yet, I'm assuming
that's why the book suggested a graphing utility.
On Sun, Feb 10, 2013 at 12:34 PM, Seth Archer
<learc83@gmail.com<mailto:learc83@gmail.com>> wrote:
>49. lim x>1 of x^2 + 1/(sqrt 2x+2) - 2
The answer in the back of the book is wrong. The limit is infinity, so
DNE. If you plug in 1, you get 2/0. I also double checked it with my nice
calculator and it gave me infinity as well.
Still working on 57. The book answer is correct on that one, though.
On Sun, Feb 10, 2013 at 11:52 AM, Sheldon Elias Schiffer
<schiffer@gsu.edu<mailto:schiffer@gsu.edu>> wrote:
I'm free still for the library meeting, if anyone is up for it.
If we go for a virtual, I can be on Skype at 6:30 PM. My skype name is:
sheldonelias
I think it could work well on Skype. I have done so programming jam
session on Skype.
Otherwise, I had some questions about two exercises on P.63. #49 & #57. I
didn't see the factoring pattern. I know what to do once I can get rid of
the radicals, but getting there was a little mysterious. The conjugate
alone is not the way, but probably is one of the steps.
Here are the questions:
49. lim x>1 of x^2 + 1/(sqrt 2x+2) - 2
57. lim x>2 of [(sqrt 6-x) - x ] / x - 2
If anyone has the factoring trick up their sleeve, remind me. It's not
coming back so far.
Thanks,
-S
________________________________
From: gsu2211@librelist.com<mailto:gsu2211@librelist.com>
[gsu2211@librelist.com<mailto:gsu2211@librelist.com>] on behalf of Seth
Archer [learc83@gmail.com<mailto:learc83@gmail.com>]
Sent: Sunday, February 10, 2013 11:30 AM
To: gsu2211@librelist.com<mailto:gsu2211@librelist.com>
Subject: [gsu2211] Sunday
Since almost everyone is having a hard time making it tonight. Why don't
we just have a virtual study group.
Anyone who has any remaining questions about anything, just post them to
the mailing list.
-Seth
```

- From:
- Lauren Peyton
- Date:
- 2013-02-11 @ 00:27

Hey guys! Sorry, I had to work late and the weather is terrible so I won't be coming down. I have a break tomorrow from 3-5:30 if anyone wants to meet up and review! On Feb 10, 2013, at 4:26 PM, Sheldon Elias Schiffer <schiffer@gsu.edu> wrote: > So it's for real nobody can come to GSU to study? Let me know if someone is coming. I can be there in 15 mins. > > -S > From: gsu2211@librelist.com [gsu2211@librelist.com] on behalf of Sheldon Elias Schiffer [schiffer@gsu.edu] > Sent: Sunday, February 10, 2013 3:27 PM > To: gsu2211@librelist.com > Subject: RE: [gsu2211] Sunday > > Ok. How about P. 72 #33. > > I started to try it, but I don't know how solve f(c)=L+epsilon where f(c) for (sin3c)/c = 3.25 > > The first of the two given epsilons is .25 > > My first move was to do the usual: > > Solve for c. My guess is that since sin oscillates, there is not one solution, but a solution within each rotation of the unit circle, so you can set the equation so that sin3c = 0 or 1, and solve for c. Am I on to something? > > 29 and 31 were easy. But bring in the trig functions with the pure algebra we did not do in class. > > -Sheldon > > From: gsu2211@librelist.com [gsu2211@librelist.com] on behalf of Seth Archer [learc83@gmail.com] > Sent: Sunday, February 10, 2013 12:55 PM > To: gsu2211@librelist.com > Subject: Re: [gsu2211] Sunday > > I'm about 99% sure there is no way to do 57 algebraically. Looks like you have to use more advanced methods we haven't learned yet, I'm assuming that's why the book suggested a graphing utility. > > > On Sun, Feb 10, 2013 at 12:34 PM, Seth Archer<learc83@gmail.com> wrote: > >49. lim x>1 of x^2 + 1/(sqrt 2x+2) - 2 > > The answer in the back of the book is wrong. The limit is infinity, so DNE. If you plug in 1, you get 2/0. I also double checked it with my nice calculator and it gave me infinity as well. > > Still working on 57. The book answer is correct on that one, though. > > > On Sun, Feb 10, 2013 at 11:52 AM, Sheldon Elias Schiffer<schiffer@gsu.edu> wrote: > I'm free still for the library meeting, if anyone is up for it. > > If we go for a virtual, I can be on Skype at 6:30 PM. My skype name is: sheldonelias > > I think it could work well on Skype. I have done so programming jam session on Skype. > > Otherwise, I had some questions about two exercises on P.63. #49 & #57. I didn't see the factoring pattern. I know what to do once I can get rid of the radicals, but getting there was a little mysterious. The conjugate alone is not the way, but probably is one of the steps. > > Here are the questions: > > 49. lim x>1 of x^2 + 1/(sqrt 2x+2) - 2 > > 57. lim x>2 of [(sqrt 6-x) - x ] / x - 2 > > If anyone has the factoring trick up their sleeve, remind me. It's not coming back so far. > > Thanks, > > -S > From: gsu2211@librelist.com[gsu2211@librelist.com] on behalf of Seth Archer [learc83@gmail.com] > Sent: Sunday, February 10, 2013 11:30 AM > To: gsu2211@librelist.com > Subject: [gsu2211] Sunday > > Since almost everyone is having a hard time making it tonight. Why don't we just have a virtual study group. > > Anyone who has any remaining questions about anything, just post them to the mailing list. > > -Seth

- From:
- Sheldon Elias Schiffer
- Date:
- 2013-02-11 @ 02:38

I get the idea, but the answers don't make sense. Where are the new functions being defined to yield a 1 and a 6 to make 1/6 (#41) or a 1 and a 3 to make 1/3 (#43)? -S ________________________________ From: gsu2211@librelist.com [gsu2211@librelist.com] on behalf of Seth Archer [learc83@gmail.com] Sent: Sunday, February 10, 2013 9:00 PM To: gsu2211@librelist.com Subject: Re: [gsu2211] Sunday They took the limit at 5. Then if you set f(5) to be equal to the limit at 5, the function becomes continuous at f(5). However we haven't learned how to take limits from expressions like that. The only thing you could do using what we've learned is to plug in numbers close to 5. On Sun, Feb 10, 2013 at 8:19 PM, Sheldon Elias Schiffer <schiffer@gsu.edu<mailto:schiffer@gsu.edu>> wrote: Seth, or anyone else, do you have a strategy for P. 90 #41 & 43? I get the idea of composite limits, but the question is, make the function continuous at 5. The answer is 1/6 and 1/3? 41. f(x)= [(sqrt (x+4)) - 3] / (x - 5) 43. f(x) = [sqrt(2x - 1) - 3] / (x - 5) Any idea how the book got that answer? -Sheldon

- From:
- Seth Archer
- Date:
- 2013-02-11 @ 03:30

They are just redefining the function into a piecewise function so the new definition is f(x) = [(sqrt (x+4)) - 3] / (x - 5), for x not equal to 5, and 1/6 for x equal to 5. On Sun, Feb 10, 2013 at 9:38 PM, Sheldon Elias Schiffer <schiffer@gsu.edu>wrote: > I get the idea, but the answers don't make sense. Where are the new > functions being defined to yield a 1 and a 6 to make 1/6 (#41) or a 1 and a > 3 to make 1/3 (#43)? > > -S > ------------------------------ > *From:* gsu2211@librelist.com [gsu2211@librelist.com] on behalf of Seth > Archer [learc83@gmail.com] > *Sent:* Sunday, February 10, 2013 9:00 PM > > *To:* gsu2211@librelist.com > *Subject:* Re: [gsu2211] Sunday > > They took the limit at 5. Then if you set f(5) to be equal to the limit > at 5, the function becomes continuous at f(5). However we haven't learned > how to take limits from expressions like that. The only thing you could do > using what we've learned is to plug in numbers close to 5. > > On Sun, Feb 10, 2013 at 8:19 PM, Sheldon Elias Schiffer <schiffer@gsu.edu>wrote: > >> Seth, or anyone else, do you have a strategy for P. 90 #41 & 43? >> >> I get the idea of composite limits, but the question is, make the >> function continuous at 5. The answer is 1/6 and 1/3? >> >> 41. f(x)= [(sqrt (x+4)) - 3] / (x - 5) >> >> 43. f(x) = [sqrt(2x - 1) - 3] / (x - 5) >> >> Any idea how the book got that answer? >> >> -Sheldon >> > >

- From:
- Seth Archer
- Date:
- 2013-02-10 @ 17:34

>49. lim x>1 of x^2 + 1/(sqrt 2x+2) - 2 The answer in the back of the book is wrong. The limit is infinity, so DNE. If you plug in 1, you get 2/0. I also double checked it with my nice calculator and it gave me infinity as well. Still working on 57. The book answer is correct on that one, though. On Sun, Feb 10, 2013 at 11:52 AM, Sheldon Elias Schiffer <schiffer@gsu.edu>wrote: > I'm free still for the library meeting, if anyone is up for it. > > If we go for a virtual, I can be on Skype at 6:30 PM. My skype name is: > sheldonelias > > I think it could work well on Skype. I have done so programming jam > session on Skype. > > Otherwise, I had some questions about two exercises on P.63. #49 & #57. > I didn't see the factoring pattern. I know what to do once I can get rid of > the radicals, but getting there was a little mysterious. The conjugate > alone is not the way, but probably is one of the steps. > > Here are the questions: > > 49. lim x>1 of x^2 + 1/(sqrt 2x+2) - 2 > > 57. lim x>2 of [(sqrt 6-x) - x ] / x - 2 > > If anyone has the factoring trick up their sleeve, remind me. It's not > coming back so far. > > Thanks, > > -S > ------------------------------ > *From:* gsu2211@librelist.com [gsu2211@librelist.com] on behalf of Seth > Archer [learc83@gmail.com] > *Sent:* Sunday, February 10, 2013 11:30 AM > *To:* gsu2211@librelist.com > *Subject:* [gsu2211] Sunday > > Since almost everyone is having a hard time making it tonight. Why > don't we just have a virtual study group. > > Anyone who has any remaining questions about anything, just post them to > the mailing list. > > -Seth >

- From:
- Seth Archer
- Date:
- 2013-02-10 @ 17:55

I'm about 99% sure there is no way to do 57 algebraically. Looks like you have to use more advanced methods we haven't learned yet, I'm assuming that's why the book suggested a graphing utility. On Sun, Feb 10, 2013 at 12:34 PM, Seth Archer <learc83@gmail.com> wrote: > >49. lim x>1 of x^2 + 1/(sqrt 2x+2) - 2 > > The answer in the back of the book is wrong. The limit is infinity, so > DNE. If you plug in 1, you get 2/0. I also double checked it with my nice > calculator and it gave me infinity as well. > > Still working on 57. The book answer is correct on that one, though. > > > On Sun, Feb 10, 2013 at 11:52 AM, Sheldon Elias Schiffer <schiffer@gsu.edu > > wrote: > >> I'm free still for the library meeting, if anyone is up for it. >> >> If we go for a virtual, I can be on Skype at 6:30 PM. My skype name is: >> sheldonelias >> >> I think it could work well on Skype. I have done so programming jam >> session on Skype. >> >> Otherwise, I had some questions about two exercises on P.63. #49 & #57. >> I didn't see the factoring pattern. I know what to do once I can get rid of >> the radicals, but getting there was a little mysterious. The conjugate >> alone is not the way, but probably is one of the steps. >> >> Here are the questions: >> >> 49. lim x>1 of x^2 + 1/(sqrt 2x+2) - 2 >> >> 57. lim x>2 of [(sqrt 6-x) - x ] / x - 2 >> >> If anyone has the factoring trick up their sleeve, remind me. It's not >> coming back so far. >> >> Thanks, >> >> -S >> ------------------------------ >> *From:* gsu2211@librelist.com [gsu2211@librelist.com] on behalf of Seth >> Archer [learc83@gmail.com] >> *Sent:* Sunday, February 10, 2013 11:30 AM >> *To:* gsu2211@librelist.com >> *Subject:* [gsu2211] Sunday >> >> Since almost everyone is having a hard time making it tonight. Why >> don't we just have a virtual study group. >> >> Anyone who has any remaining questions about anything, just post them >> to the mailing list. >> >> -Seth >> > >