Showing that $y = phi over phi -pi cdot sinphi$ tends to $-pi$ as $phi to pi$ without using limits [on hold]

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

I'm trying to manually plot a parametric function:

$$
r=fracphiphi - pi
$$

I've translated the above to Сartesian coordinate system and found zeros for $x$ and $y$ axes.

$$
cases
x = rcosphi = fracphiphi - picdot cosphi\
y = rsinphi = fracphiphi - picdot sinphi

$$

Then i wanted to find the value of $x$ and $y$ as $phi to pi$. It's clear that $x$ tends to $pm infty$ depending from which side $phi$ approaches $pi$ since $cospi = -1$. But it's not clear how to deal with the value of $y$ since it's $y =pm inftycdot0$.

I've found the limit using L'opitals rule, but the problem is taken from a precalculus section of the book i'm solving. I am not supposed to use limits or derivatives.

I've already spent a couple of hours pondering on that, but that doesn't seem very obvious to me.

So now I'm wondering how could I find that limit in a precalculus manner?

Below is the graph in case someone is curious what it's really about:

algebra-precalculus graphing-functions

share|cite|improve this question

asked Aug 2 at 16:44

roman

3941312

put on hold as unclear what you're asking by Xander Henderson, Shailesh, Mostafa Ayaz, Lord Shark the Unknown, John B Aug 3 at 9:28

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You want to compute a limit without using limits? I am very confused about what it is that you are trying to accomplish...
  – Xander Henderson
  Aug 2 at 16:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @XanderHenderson I need to know a value which is approached by $y$ as $phi to pi$, otherwise i can not build the plot.
  – roman
  Aug 2 at 16:57
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Yes, but that involves computing a limit! Indeed, the answer that you have accepted requires computation of a limit. I don't understand what you are asking...
  – Xander Henderson
  Aug 2 at 17:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @XanderHenderson I've accepted the below, because $sinx over x = 1$ can be shown geometrically without involving derivatives, and that's what i actually needed. I understand it's confusing to see "find a limit without a limit", but basically i intended to say "find a value of $y$" which is know as limit in calculus
  – roman
  Aug 2 at 17:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It can be shown without derivatives, but it cannot be shown without limits, which is why I am confused.
  – Xander Henderson
  Aug 2 at 18:09
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

I'm trying to manually plot a parametric function:

$$
r=fracphiphi - pi
$$

I've translated the above to Сartesian coordinate system and found zeros for $x$ and $y$ axes.

$$
cases
x = rcosphi = fracphiphi - picdot cosphi\
y = rsinphi = fracphiphi - picdot sinphi

$$

Then i wanted to find the value of $x$ and $y$ as $phi to pi$. It's clear that $x$ tends to $pm infty$ depending from which side $phi$ approaches $pi$ since $cospi = -1$. But it's not clear how to deal with the value of $y$ since it's $y =pm inftycdot0$.

I've found the limit using L'opitals rule, but the problem is taken from a precalculus section of the book i'm solving. I am not supposed to use limits or derivatives.

I've already spent a couple of hours pondering on that, but that doesn't seem very obvious to me.

So now I'm wondering how could I find that limit in a precalculus manner?

Below is the graph in case someone is curious what it's really about:

algebra-precalculus graphing-functions

share|cite|improve this question

asked Aug 2 at 16:44

roman

3941312

put on hold as unclear what you're asking by Xander Henderson, Shailesh, Mostafa Ayaz, Lord Shark the Unknown, John B Aug 3 at 9:28

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You want to compute a limit without using limits? I am very confused about what it is that you are trying to accomplish...
  – Xander Henderson
  Aug 2 at 16:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @XanderHenderson I need to know a value which is approached by $y$ as $phi to pi$, otherwise i can not build the plot.
  – roman
  Aug 2 at 16:57
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Yes, but that involves computing a limit! Indeed, the answer that you have accepted requires computation of a limit. I don't understand what you are asking...
  – Xander Henderson
  Aug 2 at 17:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @XanderHenderson I've accepted the below, because $sinx over x = 1$ can be shown geometrically without involving derivatives, and that's what i actually needed. I understand it's confusing to see "find a limit without a limit", but basically i intended to say "find a value of $y$" which is know as limit in calculus
  – roman
  Aug 2 at 17:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It can be shown without derivatives, but it cannot be shown without limits, which is why I am confused.
  – Xander Henderson
  Aug 2 at 18:09
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I'm trying to manually plot a parametric function:

$$
r=fracphiphi - pi
$$

I've translated the above to Сartesian coordinate system and found zeros for $x$ and $y$ axes.

$$
cases
x = rcosphi = fracphiphi - picdot cosphi\
y = rsinphi = fracphiphi - picdot sinphi

$$

Then i wanted to find the value of $x$ and $y$ as $phi to pi$. It's clear that $x$ tends to $pm infty$ depending from which side $phi$ approaches $pi$ since $cospi = -1$. But it's not clear how to deal with the value of $y$ since it's $y =pm inftycdot0$.

I've found the limit using L'opitals rule, but the problem is taken from a precalculus section of the book i'm solving. I am not supposed to use limits or derivatives.

I've already spent a couple of hours pondering on that, but that doesn't seem very obvious to me.

So now I'm wondering how could I find that limit in a precalculus manner?

Below is the graph in case someone is curious what it's really about:

algebra-precalculus graphing-functions

share|cite|improve this question

asked Aug 2 at 16:44

roman

3941312

I'm trying to manually plot a parametric function:

$$
r=fracphiphi - pi
$$

I've translated the above to Сartesian coordinate system and found zeros for $x$ and $y$ axes.

$$
cases
x = rcosphi = fracphiphi - picdot cosphi\
y = rsinphi = fracphiphi - picdot sinphi

$$

Then i wanted to find the value of $x$ and $y$ as $phi to pi$. It's clear that $x$ tends to $pm infty$ depending from which side $phi$ approaches $pi$ since $cospi = -1$. But it's not clear how to deal with the value of $y$ since it's $y =pm inftycdot0$.

I've found the limit using L'opitals rule, but the problem is taken from a precalculus section of the book i'm solving. I am not supposed to use limits or derivatives.

I've already spent a couple of hours pondering on that, but that doesn't seem very obvious to me.

So now I'm wondering how could I find that limit in a precalculus manner?

Below is the graph in case someone is curious what it's really about:

algebra-precalculus graphing-functions

share|cite|improve this question

asked Aug 2 at 16:44

roman

3941312

share|cite|improve this question

share|cite|improve this question

asked Aug 2 at 16:44

roman

3941312

asked Aug 2 at 16:44

roman

3941312

asked Aug 2 at 16:44

roman

3941312

3941312

put on hold as unclear what you're asking by Xander Henderson, Shailesh, Mostafa Ayaz, Lord Shark the Unknown, John B Aug 3 at 9:28

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

put on hold as unclear what you're asking by Xander Henderson, Shailesh, Mostafa Ayaz, Lord Shark the Unknown, John B Aug 3 at 9:28

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You want to compute a limit without using limits? I am very confused about what it is that you are trying to accomplish...
  – Xander Henderson
  Aug 2 at 16:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @XanderHenderson I need to know a value which is approached by $y$ as $phi to pi$, otherwise i can not build the plot.
  – roman
  Aug 2 at 16:57
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Yes, but that involves computing a limit! Indeed, the answer that you have accepted requires computation of a limit. I don't understand what you are asking...
  – Xander Henderson
  Aug 2 at 17:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @XanderHenderson I've accepted the below, because $sinx over x = 1$ can be shown geometrically without involving derivatives, and that's what i actually needed. I understand it's confusing to see "find a limit without a limit", but basically i intended to say "find a value of $y$" which is know as limit in calculus
  – roman
  Aug 2 at 17:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It can be shown without derivatives, but it cannot be shown without limits, which is why I am confused.
  – Xander Henderson
  Aug 2 at 18:09
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You want to compute a limit without using limits? I am very confused about what it is that you are trying to accomplish...
  – Xander Henderson
  Aug 2 at 16:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @XanderHenderson I need to know a value which is approached by $y$ as $phi to pi$, otherwise i can not build the plot.
  – roman
  Aug 2 at 16:57
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Yes, but that involves computing a limit! Indeed, the answer that you have accepted requires computation of a limit. I don't understand what you are asking...
  – Xander Henderson
  Aug 2 at 17:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @XanderHenderson I've accepted the below, because $sinx over x = 1$ can be shown geometrically without involving derivatives, and that's what i actually needed. I understand it's confusing to see "find a limit without a limit", but basically i intended to say "find a value of $y$" which is know as limit in calculus
  – roman
  Aug 2 at 17:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It can be shown without derivatives, but it cannot be shown without limits, which is why I am confused.
  – Xander Henderson
  Aug 2 at 18:09
  

  
  
  
  

3

3

You want to compute a limit without using limits? I am very confused about what it is that you are trying to accomplish...
– Xander Henderson
Aug 2 at 16:51

You want to compute a limit without using limits? I am very confused about what it is that you are trying to accomplish...
– Xander Henderson
Aug 2 at 16:51

1

1

@XanderHenderson I need to know a value which is approached by $y$ as $phi to pi$, otherwise i can not build the plot.
– roman
Aug 2 at 16:57

@XanderHenderson I need to know a value which is approached by $y$ as $phi to pi$, otherwise i can not build the plot.
– roman
Aug 2 at 16:57

2

2

Yes, but that involves computing a limit! Indeed, the answer that you have accepted requires computation of a limit. I don't understand what you are asking...
– Xander Henderson
Aug 2 at 17:01

Yes, but that involves computing a limit! Indeed, the answer that you have accepted requires computation of a limit. I don't understand what you are asking...
– Xander Henderson
Aug 2 at 17:01

1

1

@XanderHenderson I've accepted the below, because $sinx over x = 1$ can be shown geometrically without involving derivatives, and that's what i actually needed. I understand it's confusing to see "find a limit without a limit", but basically i intended to say "find a value of $y$" which is know as limit in calculus
– roman
Aug 2 at 17:10

@XanderHenderson I've accepted the below, because $sinx over x = 1$ can be shown geometrically without involving derivatives, and that's what i actually needed. I understand it's confusing to see "find a limit without a limit", but basically i intended to say "find a value of $y$" which is know as limit in calculus
– roman
Aug 2 at 17:10

It can be shown without derivatives, but it cannot be shown without limits, which is why I am confused.
– Xander Henderson
Aug 2 at 18:09

It can be shown without derivatives, but it cannot be shown without limits, which is why I am confused.
– Xander Henderson
Aug 2 at 18:09

add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

We can make use of the well-known fact that $lim_x to 0 fracsin xx = 1$ and how it is proven. The main step in the prove is to note that $cos x le fracsin xx le 1$. We will use this idea.

Now note that $sin( phi - pi) = - sin phi $ and so we have:

$$ - phi le -fracphiphi-pi sin(phi -pi) le - phi cos(phi - pi)$$

Now you can see that the middle is bounded by $-pi$ from both sides as we get closer and closer to $pi$. Hence we can conclude that $y=-pi$, as $phi to pi$

share|cite|improve this answer

answered Aug 2 at 16:56

Stefan4024

27.7k52974

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  ..and this is just using limits, of course. Just as the comment by Xander said. Very weird question...
  – DonAntonio
  Aug 2 at 17:02
  

  
  
  
  

add a comment | 

up vote
1
down vote

The problem you describe is the very the definition of a limit.

$y(pi)$ is not defined, but $y$ is defined in the neighborhood of $pi$ and $y(phi)$ is close to $pi$ when $phi$ is close to $pi.$ That is a limit!

As this is pre-calc, and you don't have a formal definition of a limit, you are going to have to hand-waive to some degree. But, you don't need L'Hopitals rule.

You can evaluate $frac sin (phi)pi - phi$ via the squeeze theorem.

Probably easier to start with $frac sin xx$

$|sin x| le |x| le |tan x|$

And $|sin (pi - phi)| le |pi - phi| le |tan (pi - phi)|$

Geometrically:

The area of the large triangle is $frac 12 tan (pi - phi)$
The area of the section of the circle is $frac 12 (pi - phi)$
And the area of the smaller traingle is $frac 12 sin (pi - phi)$

And with a little algebra:

$cos (pi - phi)| le frac sin(pi - phi)pi - phi le 1$

and $phi cos (pi - phi)| le frac phipi - phisin (pi-phi) le phi$

Establishing an upper and lower bound, each close to $pi$ as $phi$ is close to $pi$

share|cite|improve this answer

answered Aug 2 at 17:19

Doug M

39k31749

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

We can make use of the well-known fact that $lim_x to 0 fracsin xx = 1$ and how it is proven. The main step in the prove is to note that $cos x le fracsin xx le 1$. We will use this idea.

Now note that $sin( phi - pi) = - sin phi $ and so we have:

$$ - phi le -fracphiphi-pi sin(phi -pi) le - phi cos(phi - pi)$$

Now you can see that the middle is bounded by $-pi$ from both sides as we get closer and closer to $pi$. Hence we can conclude that $y=-pi$, as $phi to pi$

share|cite|improve this answer

answered Aug 2 at 16:56

Stefan4024

27.7k52974

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  ..and this is just using limits, of course. Just as the comment by Xander said. Very weird question...
  – DonAntonio
  Aug 2 at 17:02
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

We can make use of the well-known fact that $lim_x to 0 fracsin xx = 1$ and how it is proven. The main step in the prove is to note that $cos x le fracsin xx le 1$. We will use this idea.

Now note that $sin( phi - pi) = - sin phi $ and so we have:

$$ - phi le -fracphiphi-pi sin(phi -pi) le - phi cos(phi - pi)$$

Now you can see that the middle is bounded by $-pi$ from both sides as we get closer and closer to $pi$. Hence we can conclude that $y=-pi$, as $phi to pi$

share|cite|improve this answer

answered Aug 2 at 16:56

Stefan4024

27.7k52974

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  ..and this is just using limits, of course. Just as the comment by Xander said. Very weird question...
  – DonAntonio
  Aug 2 at 17:02
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

We can make use of the well-known fact that $lim_x to 0 fracsin xx = 1$ and how it is proven. The main step in the prove is to note that $cos x le fracsin xx le 1$. We will use this idea.

Now note that $sin( phi - pi) = - sin phi $ and so we have:

$$ - phi le -fracphiphi-pi sin(phi -pi) le - phi cos(phi - pi)$$

Now you can see that the middle is bounded by $-pi$ from both sides as we get closer and closer to $pi$. Hence we can conclude that $y=-pi$, as $phi to pi$

share|cite|improve this answer

answered Aug 2 at 16:56

Stefan4024

27.7k52974

We can make use of the well-known fact that $lim_x to 0 fracsin xx = 1$ and how it is proven. The main step in the prove is to note that $cos x le fracsin xx le 1$. We will use this idea.

Now note that $sin( phi - pi) = - sin phi $ and so we have:

$$ - phi le -fracphiphi-pi sin(phi -pi) le - phi cos(phi - pi)$$

Now you can see that the middle is bounded by $-pi$ from both sides as we get closer and closer to $pi$. Hence we can conclude that $y=-pi$, as $phi to pi$

share|cite|improve this answer

answered Aug 2 at 16:56

Stefan4024

27.7k52974

share|cite|improve this answer

share|cite|improve this answer

answered Aug 2 at 16:56

Stefan4024

27.7k52974

answered Aug 2 at 16:56

Stefan4024

27.7k52974

answered Aug 2 at 16:56

Stefan4024

27.7k52974

27.7k52974

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  ..and this is just using limits, of course. Just as the comment by Xander said. Very weird question...
  – DonAntonio
  Aug 2 at 17:02
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  ..and this is just using limits, of course. Just as the comment by Xander said. Very weird question...
  – DonAntonio
  Aug 2 at 17:02
  

  
  
  
  

1

1

..and this is just using limits, of course. Just as the comment by Xander said. Very weird question...
– DonAntonio
Aug 2 at 17:02

..and this is just using limits, of course. Just as the comment by Xander said. Very weird question...
– DonAntonio
Aug 2 at 17:02

add a comment | 

up vote
1
down vote

The problem you describe is the very the definition of a limit.

$y(pi)$ is not defined, but $y$ is defined in the neighborhood of $pi$ and $y(phi)$ is close to $pi$ when $phi$ is close to $pi.$ That is a limit!

As this is pre-calc, and you don't have a formal definition of a limit, you are going to have to hand-waive to some degree. But, you don't need L'Hopitals rule.

You can evaluate $frac sin (phi)pi - phi$ via the squeeze theorem.

Probably easier to start with $frac sin xx$

$|sin x| le |x| le |tan x|$

And $|sin (pi - phi)| le |pi - phi| le |tan (pi - phi)|$

Geometrically:

The area of the large triangle is $frac 12 tan (pi - phi)$
The area of the section of the circle is $frac 12 (pi - phi)$
And the area of the smaller traingle is $frac 12 sin (pi - phi)$

And with a little algebra:

$cos (pi - phi)| le frac sin(pi - phi)pi - phi le 1$

and $phi cos (pi - phi)| le frac phipi - phisin (pi-phi) le phi$

Establishing an upper and lower bound, each close to $pi$ as $phi$ is close to $pi$

share|cite|improve this answer

answered Aug 2 at 17:19

Doug M

39k31749

add a comment | 

up vote
1
down vote

The problem you describe is the very the definition of a limit.

$y(pi)$ is not defined, but $y$ is defined in the neighborhood of $pi$ and $y(phi)$ is close to $pi$ when $phi$ is close to $pi.$ That is a limit!

As this is pre-calc, and you don't have a formal definition of a limit, you are going to have to hand-waive to some degree. But, you don't need L'Hopitals rule.

You can evaluate $frac sin (phi)pi - phi$ via the squeeze theorem.

Probably easier to start with $frac sin xx$

$|sin x| le |x| le |tan x|$

And $|sin (pi - phi)| le |pi - phi| le |tan (pi - phi)|$

Geometrically:

The area of the large triangle is $frac 12 tan (pi - phi)$
The area of the section of the circle is $frac 12 (pi - phi)$
And the area of the smaller traingle is $frac 12 sin (pi - phi)$

And with a little algebra:

$cos (pi - phi)| le frac sin(pi - phi)pi - phi le 1$

and $phi cos (pi - phi)| le frac phipi - phisin (pi-phi) le phi$

Establishing an upper and lower bound, each close to $pi$ as $phi$ is close to $pi$

share|cite|improve this answer

answered Aug 2 at 17:19

Doug M

39k31749

add a comment | 

up vote
1
down vote

up vote
1
down vote

The problem you describe is the very the definition of a limit.

$y(pi)$ is not defined, but $y$ is defined in the neighborhood of $pi$ and $y(phi)$ is close to $pi$ when $phi$ is close to $pi.$ That is a limit!

As this is pre-calc, and you don't have a formal definition of a limit, you are going to have to hand-waive to some degree. But, you don't need L'Hopitals rule.

You can evaluate $frac sin (phi)pi - phi$ via the squeeze theorem.

Probably easier to start with $frac sin xx$

$|sin x| le |x| le |tan x|$

And $|sin (pi - phi)| le |pi - phi| le |tan (pi - phi)|$

Geometrically:

The area of the large triangle is $frac 12 tan (pi - phi)$
The area of the section of the circle is $frac 12 (pi - phi)$
And the area of the smaller traingle is $frac 12 sin (pi - phi)$

And with a little algebra:

$cos (pi - phi)| le frac sin(pi - phi)pi - phi le 1$

and $phi cos (pi - phi)| le frac phipi - phisin (pi-phi) le phi$

Establishing an upper and lower bound, each close to $pi$ as $phi$ is close to $pi$

share|cite|improve this answer

answered Aug 2 at 17:19

Doug M

39k31749

The problem you describe is the very the definition of a limit.

$y(pi)$ is not defined, but $y$ is defined in the neighborhood of $pi$ and $y(phi)$ is close to $pi$ when $phi$ is close to $pi.$ That is a limit!

As this is pre-calc, and you don't have a formal definition of a limit, you are going to have to hand-waive to some degree. But, you don't need L'Hopitals rule.

You can evaluate $frac sin (phi)pi - phi$ via the squeeze theorem.

Probably easier to start with $frac sin xx$

$|sin x| le |x| le |tan x|$

And $|sin (pi - phi)| le |pi - phi| le |tan (pi - phi)|$

Geometrically:

The area of the large triangle is $frac 12 tan (pi - phi)$
The area of the section of the circle is $frac 12 (pi - phi)$
And the area of the smaller traingle is $frac 12 sin (pi - phi)$

And with a little algebra:

$cos (pi - phi)| le frac sin(pi - phi)pi - phi le 1$

and $phi cos (pi - phi)| le frac phipi - phisin (pi-phi) le phi$

Establishing an upper and lower bound, each close to $pi$ as $phi$ is close to $pi$

share|cite|improve this answer

answered Aug 2 at 17:19

Doug M

39k31749

share|cite|improve this answer

share|cite|improve this answer

answered Aug 2 at 17:19

Doug M

39k31749

answered Aug 2 at 17:19

Doug M

39k31749

answered Aug 2 at 17:19

Doug M

39k31749

39k31749

add a comment | 

add a comment |