Mixing
Complex infinity and $frac00$ indeterminants
Clash Royale CLAN TAG#URR8PPP
up vote
-1
down vote
favorite
I have been trying to determine the following two limits, Wolfram Alpha computes (1) to be equal to something it refers to as complex infinity, and (2) to be indeterminate. So I also would like to know the difference between "Complex Infinity" and an indeterminate, as well as the step by step working out for demonstrating these results.
$$lim _xrightarrow 3/2Biggl(fraclfloorln(x^3)rfloorlfloorln(x)rfloorBiggr)quadquadquadquadquadquadquadquad (1)$$
$$lim _xrightarrow 3/2Biggl(fraclfloorln(x^2)rfloorlfloorln(x)rfloorBiggr)quadquadquadquadquadquadquadquad(2)$$
As far as I can tell, (1) is an indeterminate limit of the form $frac10$ and (2) is an indeterminate of the form $frac00$, on what basis can we say that one is different to the other, or more so, what is the argument for there being a significant difference we should account for in all such cases that evaluate to the two differing "Categories" of indeterminate forms?
number-theory limits
asked Jul 17 at 16:57
Adam
13112
 |Â
show 1 more comment
up vote
-1
down vote
favorite
I have been trying to determine the following two limits, Wolfram Alpha computes (1) to be equal to something it refers to as complex infinity, and (2) to be indeterminate. So I also would like to know the difference between "Complex Infinity" and an indeterminate, as well as the step by step working out for demonstrating these results.
$$lim _xrightarrow 3/2Biggl(fraclfloorln(x^3)rfloorlfloorln(x)rfloorBiggr)quadquadquadquadquadquadquadquad (1)$$
$$lim _xrightarrow 3/2Biggl(fraclfloorln(x^2)rfloorlfloorln(x)rfloorBiggr)quadquadquadquadquadquadquadquad(2)$$
As far as I can tell, (1) is an indeterminate limit of the form $frac10$ and (2) is an indeterminate of the form $frac00$, on what basis can we say that one is different to the other, or more so, what is the argument for there being a significant difference we should account for in all such cases that evaluate to the two differing "Categories" of indeterminate forms?
number-theory limits
asked Jul 17 at 16:57
Adam
13112
1
I would say "both of them undefined".
– Holo
Jul 17 at 17:01
Agreed. But I just found it curious that their software has been coded to return a distinctly different answer so I guess that's the point of my post
– Adam
Jul 17 at 17:02
complex infinity is not widely used, but wolfram alpha use it when we have $|x|=infty$ and $arg(x)=$unknown/undefined. Undefined is when the value can be one of few possible values without way to determine which one it is. see also math.stackexchange.com/questions/1294852/…
– Holo
Jul 17 at 17:05
Sure the definition they provided for it was orientated around the programming of wolfram alpha, but I just felt as if such a definition was one with a substantial load in terms mathematically fundamental meaning, so it naturally raised an eye brow
– Adam
Jul 17 at 17:08
There are 13 Indeterminate forms, starting with 0/0.
– Ed Pegg
Jul 17 at 17:19
 |Â
show 1 more comment
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I have been trying to determine the following two limits, Wolfram Alpha computes (1) to be equal to something it refers to as complex infinity, and (2) to be indeterminate. So I also would like to know the difference between "Complex Infinity" and an indeterminate, as well as the step by step working out for demonstrating these results.
$$lim _xrightarrow 3/2Biggl(fraclfloorln(x^3)rfloorlfloorln(x)rfloorBiggr)quadquadquadquadquadquadquadquad (1)$$
$$lim _xrightarrow 3/2Biggl(fraclfloorln(x^2)rfloorlfloorln(x)rfloorBiggr)quadquadquadquadquadquadquadquad(2)$$
As far as I can tell, (1) is an indeterminate limit of the form $frac10$ and (2) is an indeterminate of the form $frac00$, on what basis can we say that one is different to the other, or more so, what is the argument for there being a significant difference we should account for in all such cases that evaluate to the two differing "Categories" of indeterminate forms?
number-theory limits
asked Jul 17 at 16:57
Adam
13112
I have been trying to determine the following two limits, Wolfram Alpha computes (1) to be equal to something it refers to as complex infinity, and (2) to be indeterminate. So I also would like to know the difference between "Complex Infinity" and an indeterminate, as well as the step by step working out for demonstrating these results.
$$lim _xrightarrow 3/2Biggl(fraclfloorln(x^3)rfloorlfloorln(x)rfloorBiggr)quadquadquadquadquadquadquadquad (1)$$
$$lim _xrightarrow 3/2Biggl(fraclfloorln(x^2)rfloorlfloorln(x)rfloorBiggr)quadquadquadquadquadquadquadquad(2)$$
As far as I can tell, (1) is an indeterminate limit of the form $frac10$ and (2) is an indeterminate of the form $frac00$, on what basis can we say that one is different to the other, or more so, what is the argument for there being a significant difference we should account for in all such cases that evaluate to the two differing "Categories" of indeterminate forms?
number-theory limits
asked Jul 17 at 16:57
Adam
13112
asked Jul 17 at 16:57
Adam
13112
asked Jul 17 at 16:57
Adam
13112
asked Jul 17 at 16:57
Adam
13112
13112
1
I would say "both of them undefined".
– Holo
Jul 17 at 17:01
Agreed. But I just found it curious that their software has been coded to return a distinctly different answer so I guess that's the point of my post
– Adam
Jul 17 at 17:02
complex infinity is not widely used, but wolfram alpha use it when we have $|x|=infty$ and $arg(x)=$unknown/undefined. Undefined is when the value can be one of few possible values without way to determine which one it is. see also math.stackexchange.com/questions/1294852/…
– Holo
Jul 17 at 17:05
Sure the definition they provided for it was orientated around the programming of wolfram alpha, but I just felt as if such a definition was one with a substantial load in terms mathematically fundamental meaning, so it naturally raised an eye brow
– Adam
Jul 17 at 17:08
There are 13 Indeterminate forms, starting with 0/0.
– Ed Pegg
Jul 17 at 17:19
 |Â
show 1 more comment
1
I would say "both of them undefined".
– Holo
Jul 17 at 17:01
Agreed. But I just found it curious that their software has been coded to return a distinctly different answer so I guess that's the point of my post
– Adam
Jul 17 at 17:02
complex infinity is not widely used, but wolfram alpha use it when we have $|x|=infty$ and $arg(x)=$unknown/undefined. Undefined is when the value can be one of few possible values without way to determine which one it is. see also math.stackexchange.com/questions/1294852/…
– Holo
Jul 17 at 17:05
Sure the definition they provided for it was orientated around the programming of wolfram alpha, but I just felt as if such a definition was one with a substantial load in terms mathematically fundamental meaning, so it naturally raised an eye brow
– Adam
Jul 17 at 17:08
There are 13 Indeterminate forms, starting with 0/0.
– Ed Pegg
Jul 17 at 17:19
1
1
I would say "both of them undefined".
– Holo
Jul 17 at 17:01
I would say "both of them undefined".
– Holo
Jul 17 at 17:01
Agreed. But I just found it curious that their software has been coded to return a distinctly different answer so I guess that's the point of my post
– Adam
Jul 17 at 17:02
Agreed. But I just found it curious that their software has been coded to return a distinctly different answer so I guess that's the point of my post
– Adam
Jul 17 at 17:02
complex infinity is not widely used, but wolfram alpha use it when we have $|x|=infty$ and $arg(x)=$unknown/undefined. Undefined is when the value can be one of few possible values without way to determine which one it is. see also math.stackexchange.com/questions/1294852/…
– Holo
Jul 17 at 17:05
complex infinity is not widely used, but wolfram alpha use it when we have $|x|=infty$ and $arg(x)=$unknown/undefined. Undefined is when the value can be one of few possible values without way to determine which one it is. see also math.stackexchange.com/questions/1294852/…
– Holo
Jul 17 at 17:05
Sure the definition they provided for it was orientated around the programming of wolfram alpha, but I just felt as if such a definition was one with a substantial load in terms mathematically fundamental meaning, so it naturally raised an eye brow
– Adam
Jul 17 at 17:08
Sure the definition they provided for it was orientated around the programming of wolfram alpha, but I just felt as if such a definition was one with a substantial load in terms mathematically fundamental meaning, so it naturally raised an eye brow
– Adam
Jul 17 at 17:08
There are 13 Indeterminate forms, starting with 0/0.
– Ed Pegg
Jul 17 at 17:19
There are 13 Indeterminate forms, starting with 0/0.
– Ed Pegg
Jul 17 at 17:19
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
1
down vote
The difference is in the limit of the top.
Note that $$ln(3/2)=0.4054...$$ so its integer part is $0$
Similarly $$ln(9/4)=0.810934054...$$ so its integer part is $0$ as well.
On the other hand $$ln(27/8)=1.21639...$$ so iteger part is $1$
That is the first limit is of $$ 0/0 $$ form while the second is of $$1/0$$ form.
answered Jul 17 at 17:24
Mohammad Riazi-Kermani
27.5k41852
1
The first limit is not $0/0$: it's not defined in a punctured neighborhood of $3/2$, so it makes no sense at all.
– egreg
Jul 17 at 17:37
@egreg you are correct in that the quotient is not defined . I was talking abot the form not the limit.
– Mohammad Riazi-Kermani
Jul 17 at 17:45
Sure ok thankyou sir, I suppose I need more elaboration in that I was aware of the difference in the numerators of the forms, more as to how this should affect the way in which we contemplate the functions and the arguments for which they evaluate to these differing forms
– Adam
Jul 17 at 17:48
@Adam Thanks for your attention .
– Mohammad Riazi-Kermani
Jul 17 at 17:50
add a comment |Â
up vote
1
down vote
Suppose a function $f$ is defined on a set $D$. In order for
$$
lim_xto af(x)
$$
to at least make sense, $a$ should be an accumulation point (aka limit point) of $Dsetminusa$. The two expressions you have are defined where $ln(x)$ exists and
$$
lfloorln(x)rfloorne0
$$
Now $lfloorln(x)rfloor=0$ if and only if $0leln(x)<1$, which is the same as $1le x<e$. Thus the functions are defined over $(0,1)cup(einfty)$.
As you see, $3/2$ is not an accumulation point of the domain of the functions, so neither limit makes sense.
Wolframalpha does strange things for assigning a meaning to undefined expressions: don't take its answers as revealed truth.
answered Jul 17 at 17:47
egreg
164k1180187
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The difference is in the limit of the top.
Note that $$ln(3/2)=0.4054...$$ so its integer part is $0$
Similarly $$ln(9/4)=0.810934054...$$ so its integer part is $0$ as well.
On the other hand $$ln(27/8)=1.21639...$$ so iteger part is $1$
That is the first limit is of $$ 0/0 $$ form while the second is of $$1/0$$ form.
answered Jul 17 at 17:24
Mohammad Riazi-Kermani
27.5k41852
1
The first limit is not $0/0$: it's not defined in a punctured neighborhood of $3/2$, so it makes no sense at all.
– egreg
Jul 17 at 17:37
@egreg you are correct in that the quotient is not defined . I was talking abot the form not the limit.
– Mohammad Riazi-Kermani
Jul 17 at 17:45
Sure ok thankyou sir, I suppose I need more elaboration in that I was aware of the difference in the numerators of the forms, more as to how this should affect the way in which we contemplate the functions and the arguments for which they evaluate to these differing forms
– Adam
Jul 17 at 17:48
@Adam Thanks for your attention .
– Mohammad Riazi-Kermani
Jul 17 at 17:50
add a comment |Â
up vote
1
down vote
The difference is in the limit of the top.
Note that $$ln(3/2)=0.4054...$$ so its integer part is $0$
Similarly $$ln(9/4)=0.810934054...$$ so its integer part is $0$ as well.
On the other hand $$ln(27/8)=1.21639...$$ so iteger part is $1$
That is the first limit is of $$ 0/0 $$ form while the second is of $$1/0$$ form.
answered Jul 17 at 17:24
Mohammad Riazi-Kermani
27.5k41852
1
The first limit is not $0/0$: it's not defined in a punctured neighborhood of $3/2$, so it makes no sense at all.
– egreg
Jul 17 at 17:37
@egreg you are correct in that the quotient is not defined . I was talking abot the form not the limit.
– Mohammad Riazi-Kermani
Jul 17 at 17:45
Sure ok thankyou sir, I suppose I need more elaboration in that I was aware of the difference in the numerators of the forms, more as to how this should affect the way in which we contemplate the functions and the arguments for which they evaluate to these differing forms
– Adam
Jul 17 at 17:48
@Adam Thanks for your attention .
– Mohammad Riazi-Kermani
Jul 17 at 17:50
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The difference is in the limit of the top.
Note that $$ln(3/2)=0.4054...$$ so its integer part is $0$
Similarly $$ln(9/4)=0.810934054...$$ so its integer part is $0$ as well.
On the other hand $$ln(27/8)=1.21639...$$ so iteger part is $1$
That is the first limit is of $$ 0/0 $$ form while the second is of $$1/0$$ form.
answered Jul 17 at 17:24
Mohammad Riazi-Kermani
27.5k41852
The difference is in the limit of the top.
Note that $$ln(3/2)=0.4054...$$ so its integer part is $0$
Similarly $$ln(9/4)=0.810934054...$$ so its integer part is $0$ as well.
On the other hand $$ln(27/8)=1.21639...$$ so iteger part is $1$
That is the first limit is of $$ 0/0 $$ form while the second is of $$1/0$$ form.
answered Jul 17 at 17:24
Mohammad Riazi-Kermani
27.5k41852
edited Jul 17 at 17:34
answered Jul 17 at 17:24
Mohammad Riazi-Kermani
27.5k41852
answered Jul 17 at 17:24
Mohammad Riazi-Kermani
27.5k41852
answered Jul 17 at 17:24
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
1
The first limit is not $0/0$: it's not defined in a punctured neighborhood of $3/2$, so it makes no sense at all.
– egreg
Jul 17 at 17:37
@egreg you are correct in that the quotient is not defined . I was talking abot the form not the limit.
– Mohammad Riazi-Kermani
Jul 17 at 17:45
Sure ok thankyou sir, I suppose I need more elaboration in that I was aware of the difference in the numerators of the forms, more as to how this should affect the way in which we contemplate the functions and the arguments for which they evaluate to these differing forms
– Adam
Jul 17 at 17:48
@Adam Thanks for your attention .
– Mohammad Riazi-Kermani
Jul 17 at 17:50
add a comment |Â
1
The first limit is not $0/0$: it's not defined in a punctured neighborhood of $3/2$, so it makes no sense at all.
– egreg
Jul 17 at 17:37
@egreg you are correct in that the quotient is not defined . I was talking abot the form not the limit.
– Mohammad Riazi-Kermani
Jul 17 at 17:45
Sure ok thankyou sir, I suppose I need more elaboration in that I was aware of the difference in the numerators of the forms, more as to how this should affect the way in which we contemplate the functions and the arguments for which they evaluate to these differing forms
– Adam
Jul 17 at 17:48
@Adam Thanks for your attention .
– Mohammad Riazi-Kermani
Jul 17 at 17:50
1
1
The first limit is not $0/0$: it's not defined in a punctured neighborhood of $3/2$, so it makes no sense at all.
– egreg
Jul 17 at 17:37
The first limit is not $0/0$: it's not defined in a punctured neighborhood of $3/2$, so it makes no sense at all.
– egreg
Jul 17 at 17:37
@egreg you are correct in that the quotient is not defined . I was talking abot the form not the limit.
– Mohammad Riazi-Kermani
Jul 17 at 17:45
@egreg you are correct in that the quotient is not defined . I was talking abot the form not the limit.
– Mohammad Riazi-Kermani
Jul 17 at 17:45
Sure ok thankyou sir, I suppose I need more elaboration in that I was aware of the difference in the numerators of the forms, more as to how this should affect the way in which we contemplate the functions and the arguments for which they evaluate to these differing forms
– Adam
Jul 17 at 17:48
Sure ok thankyou sir, I suppose I need more elaboration in that I was aware of the difference in the numerators of the forms, more as to how this should affect the way in which we contemplate the functions and the arguments for which they evaluate to these differing forms
– Adam
Jul 17 at 17:48
@Adam Thanks for your attention .
– Mohammad Riazi-Kermani
Jul 17 at 17:50
@Adam Thanks for your attention .
– Mohammad Riazi-Kermani
Jul 17 at 17:50
add a comment |Â
up vote
1
down vote
Suppose a function $f$ is defined on a set $D$. In order for
$$
lim_xto af(x)
$$
to at least make sense, $a$ should be an accumulation point (aka limit point) of $Dsetminusa$. The two expressions you have are defined where $ln(x)$ exists and
$$
lfloorln(x)rfloorne0
$$
Now $lfloorln(x)rfloor=0$ if and only if $0leln(x)<1$, which is the same as $1le x<e$. Thus the functions are defined over $(0,1)cup(einfty)$.
As you see, $3/2$ is not an accumulation point of the domain of the functions, so neither limit makes sense.
Wolframalpha does strange things for assigning a meaning to undefined expressions: don't take its answers as revealed truth.
answered Jul 17 at 17:47
egreg
164k1180187
add a comment |Â
up vote
1
down vote
Suppose a function $f$ is defined on a set $D$. In order for
$$
lim_xto af(x)
$$
to at least make sense, $a$ should be an accumulation point (aka limit point) of $Dsetminusa$. The two expressions you have are defined where $ln(x)$ exists and
$$
lfloorln(x)rfloorne0
$$
Now $lfloorln(x)rfloor=0$ if and only if $0leln(x)<1$, which is the same as $1le x<e$. Thus the functions are defined over $(0,1)cup(einfty)$.
As you see, $3/2$ is not an accumulation point of the domain of the functions, so neither limit makes sense.
Wolframalpha does strange things for assigning a meaning to undefined expressions: don't take its answers as revealed truth.
answered Jul 17 at 17:47
egreg
164k1180187
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Suppose a function $f$ is defined on a set $D$. In order for
$$
lim_xto af(x)
$$
to at least make sense, $a$ should be an accumulation point (aka limit point) of $Dsetminusa$. The two expressions you have are defined where $ln(x)$ exists and
$$
lfloorln(x)rfloorne0
$$
Now $lfloorln(x)rfloor=0$ if and only if $0leln(x)<1$, which is the same as $1le x<e$. Thus the functions are defined over $(0,1)cup(einfty)$.
As you see, $3/2$ is not an accumulation point of the domain of the functions, so neither limit makes sense.
Wolframalpha does strange things for assigning a meaning to undefined expressions: don't take its answers as revealed truth.
answered Jul 17 at 17:47
egreg
164k1180187
Suppose a function $f$ is defined on a set $D$. In order for
$$
lim_xto af(x)
$$
to at least make sense, $a$ should be an accumulation point (aka limit point) of $Dsetminusa$. The two expressions you have are defined where $ln(x)$ exists and
$$
lfloorln(x)rfloorne0
$$
Now $lfloorln(x)rfloor=0$ if and only if $0leln(x)<1$, which is the same as $1le x<e$. Thus the functions are defined over $(0,1)cup(einfty)$.
As you see, $3/2$ is not an accumulation point of the domain of the functions, so neither limit makes sense.
Wolframalpha does strange things for assigning a meaning to undefined expressions: don't take its answers as revealed truth.
answered Jul 17 at 17:47
egreg
164k1180187
answered Jul 17 at 17:47
egreg
164k1180187
answered Jul 17 at 17:47
egreg
164k1180187
answered Jul 17 at 17:47
egreg
164k1180187
164k1180187
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2854710%2fcomplex-infinity-and-frac00-indeterminants%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
I would say "both of them undefined".
– Holo
Jul 17 at 17:01
Agreed. But I just found it curious that their software has been coded to return a distinctly different answer so I guess that's the point of my post
– Adam
Jul 17 at 17:02
complex infinity is not widely used, but wolfram alpha use it when we have $|x|=infty$ and $arg(x)=$unknown/undefined. Undefined is when the value can be one of few possible values without way to determine which one it is. see also math.stackexchange.com/questions/1294852/…
– Holo
Jul 17 at 17:05
Sure the definition they provided for it was orientated around the programming of wolfram alpha, but I just felt as if such a definition was one with a substantial load in terms mathematically fundamental meaning, so it naturally raised an eye brow
– Adam
Jul 17 at 17:08
There are 13 Indeterminate forms, starting with 0/0.
– Ed Pegg
Jul 17 at 17:19