Is $sqrt[0]0$ defined? [closed]

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Is $sqrt[0]0$ defined?

How about the limit? $lim_xto 0x^frac1x$

Is $sqrt[0]n,,,,nneq0$ any different?

algebra-precalculus

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asked Jul 15 at 10:37

Mitchell Browne

158310

closed as unclear what you're asking by Alex Francisco, Trần Thúc Minh Trí, José Carlos Santos, Ethan Bolker, user190080 Jul 15 at 21:00

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
  

  
  
  
  
  
  

  
  $sqrt[0]n$ for $nne 1$ has the noteworthy issue of not making sense in the first place: there is no real, complex of whatever number $alpha$ such that $alpha^0=26$.
  – Saucy O'Path
  Jul 15 at 11:03
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Please define $sqrt[0]n$. Do you really mean $n^0$? And are you really talking about $0^0$?
  – Michael
  Jul 15 at 11:06
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  This must be a duplicate
  – Jam
  Jul 15 at 11:15
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

Is $sqrt[0]0$ defined?

How about the limit? $lim_xto 0x^frac1x$

Is $sqrt[0]n,,,,nneq0$ any different?

algebra-precalculus

share|cite|improve this question

asked Jul 15 at 10:37

Mitchell Browne

158310

closed as unclear what you're asking by Alex Francisco, Trần Thúc Minh Trí, José Carlos Santos, Ethan Bolker, user190080 Jul 15 at 21:00

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
  

  
  
  
  
  
  

  
  $sqrt[0]n$ for $nne 1$ has the noteworthy issue of not making sense in the first place: there is no real, complex of whatever number $alpha$ such that $alpha^0=26$.
  – Saucy O'Path
  Jul 15 at 11:03
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Please define $sqrt[0]n$. Do you really mean $n^0$? And are you really talking about $0^0$?
  – Michael
  Jul 15 at 11:06
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  This must be a duplicate
  – Jam
  Jul 15 at 11:15
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Is $sqrt[0]0$ defined?

How about the limit? $lim_xto 0x^frac1x$

Is $sqrt[0]n,,,,nneq0$ any different?

algebra-precalculus

share|cite|improve this question

asked Jul 15 at 10:37

Mitchell Browne

158310

Is $sqrt[0]0$ defined?

How about the limit? $lim_xto 0x^frac1x$

Is $sqrt[0]n,,,,nneq0$ any different?

algebra-precalculus

share|cite|improve this question

asked Jul 15 at 10:37

Mitchell Browne

158310

share|cite|improve this question

share|cite|improve this question

asked Jul 15 at 10:37

Mitchell Browne

158310

asked Jul 15 at 10:37

Mitchell Browne

158310

asked Jul 15 at 10:37

Mitchell Browne

158310

158310

closed as unclear what you're asking by Alex Francisco, Trần Thúc Minh Trí, José Carlos Santos, Ethan Bolker, user190080 Jul 15 at 21:00

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.

closed as unclear what you're asking by Alex Francisco, Trần Thúc Minh Trí, José Carlos Santos, Ethan Bolker, user190080 Jul 15 at 21:00

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
  

  
  
  
  
  
  

  
  $sqrt[0]n$ for $nne 1$ has the noteworthy issue of not making sense in the first place: there is no real, complex of whatever number $alpha$ such that $alpha^0=26$.
  – Saucy O'Path
  Jul 15 at 11:03
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Please define $sqrt[0]n$. Do you really mean $n^0$? And are you really talking about $0^0$?
  – Michael
  Jul 15 at 11:06
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  This must be a duplicate
  – Jam
  Jul 15 at 11:15
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $sqrt[0]n$ for $nne 1$ has the noteworthy issue of not making sense in the first place: there is no real, complex of whatever number $alpha$ such that $alpha^0=26$.
  – Saucy O'Path
  Jul 15 at 11:03
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Please define $sqrt[0]n$. Do you really mean $n^0$? And are you really talking about $0^0$?
  – Michael
  Jul 15 at 11:06
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  This must be a duplicate
  – Jam
  Jul 15 at 11:15
  

  
  
  
  

3

3

$sqrt[0]n$ for $nne 1$ has the noteworthy issue of not making sense in the first place: there is no real, complex of whatever number $alpha$ such that $alpha^0=26$.
– Saucy O'Path
Jul 15 at 11:03

$sqrt[0]n$ for $nne 1$ has the noteworthy issue of not making sense in the first place: there is no real, complex of whatever number $alpha$ such that $alpha^0=26$.
– Saucy O'Path
Jul 15 at 11:03

3

3

Please define $sqrt[0]n$. Do you really mean $n^0$? And are you really talking about $0^0$?
– Michael
Jul 15 at 11:06

Please define $sqrt[0]n$. Do you really mean $n^0$? And are you really talking about $0^0$?
– Michael
Jul 15 at 11:06

2

2

This must be a duplicate
– Jam
Jul 15 at 11:15

This must be a duplicate
– Jam
Jul 15 at 11:15

add a comment | 

4 Answers
4

active

oldest

votes

up vote
5
down vote



accepted

We have $$lim_xto0^+x^1/x=0$$ since $x^1/x$ is not defined for all negative $x$.

This is because as $xto0$, $frac1xtoinfty$ and multiplying a number $<1$ by itself $n$ times (as $ntoinfty$) gets you to zero.

Notice that we are writing $xto0$, not $x=0$, as $x^1/x=0^infty$ is indeterminate.

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edited Jul 15 at 11:14

answered Jul 15 at 10:41

TheSimpliFire

9,69261951

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I hope that $x>0$ as you take your limit...?
  – Michael
  Jul 15 at 11:09
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Yes, as going from the negative side isn't defined.
  – TheSimpliFire
  Jul 15 at 11:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Perhaps it is better to write $lim_xrightarrow 0^+$.
  – Michael
  Jul 15 at 11:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Michael Edited.
  – TheSimpliFire
  Jul 15 at 11:14
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

The limit dose not exist because for negative values of $x$, $x^1/x$ is not defined.

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edited Jul 15 at 11:08

answered Jul 15 at 10:55

Mohammad Riazi-Kermani

27.6k41852

add a comment | 

up vote
1
down vote

Let $L = lim_x to 0rootxof x$. Then, taking logarithm on both sides,

$$beginaligned
ln L = lim_x to 0fraclnxx
endaligned$$

This quantity is not defined, that is, it is an indeterminate of the form $fracinfty0$.

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edited Jul 15 at 11:22

answered Jul 15 at 10:46

Quasar

697412

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Have you tried L'Hopital?
  – TheSimpliFire
  Jul 15 at 11:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  As far as I remember, L' Hopital is applicable only when the $lim_x to cf(x) = lim_x to c g(x) =0$ or $pm infty$. So, it wouldn't be applicable in this case.
  – Quasar
  Jul 15 at 11:19
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

No. It is undefined, and here's why.

^{You write $sqrt[0]n$, but I will write it as $sqrt[0]x$ with respect to the limit (because it has a variable $x$).}

Regarding the zeroth root,$$beginaligntextLet ;;,y&=sqrt[0]x, \ textthen ;y^0&=x.endalign$$

Since anything raised to the power of $0$ is $1$, it follows then that $x=1$. $$therefore sqrt[0]xtext is undefinedLeftrightarrow xneq 1.tag$therefore sqrt[0]0$ is undefined.$$ But $y$ could be any number, too, so this means $sqrt[0]x$ has no definite value, even if $xneq 1$.

$$therefore sqrt[0]xtext is undefined forall x.$$

---

Regarding the limit,
$$Lambda=lim_xto0x^frac 1x$$ this is of course equal to $0$. $overbracetextSince $0^n=0$ for all $nneq 0$,^large Statement (1)$ then as $xto 0$, it follows that $x^frac 1xto 0$, and by definition of a limit, we get that $Lambda = 0$.

But this does not mean that $1div 0$ is defined — it is not; division by zero is undefined, regardless of the value of the numerator. We just have to consider the limit, $$beginalignLambda_0=lim_xto0frac 1x&=infty, \ because lim_xtoinftyfrac 1x&=0.endalign$$ Therefore, when evaluating the limit $Lambda$, we notice that $1div xto infty$.

And thus, by Statement $(1)$, it is fully clear that $Lambda = 0$; id est, $$lim_xto 0frac 1x=0.$$

---

If you are still unsure and/or confused, take a look at the best answer (in my opinion).

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edited Jul 15 at 11:53

answered Jul 15 at 11:36

user477343

4,21731039

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote



accepted

We have $$lim_xto0^+x^1/x=0$$ since $x^1/x$ is not defined for all negative $x$.

This is because as $xto0$, $frac1xtoinfty$ and multiplying a number $<1$ by itself $n$ times (as $ntoinfty$) gets you to zero.

Notice that we are writing $xto0$, not $x=0$, as $x^1/x=0^infty$ is indeterminate.

share|cite|improve this answer

edited Jul 15 at 11:14

answered Jul 15 at 10:41

TheSimpliFire

9,69261951

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I hope that $x>0$ as you take your limit...?
  – Michael
  Jul 15 at 11:09
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Yes, as going from the negative side isn't defined.
  – TheSimpliFire
  Jul 15 at 11:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Perhaps it is better to write $lim_xrightarrow 0^+$.
  – Michael
  Jul 15 at 11:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Michael Edited.
  – TheSimpliFire
  Jul 15 at 11:14
  
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

We have $$lim_xto0^+x^1/x=0$$ since $x^1/x$ is not defined for all negative $x$.

This is because as $xto0$, $frac1xtoinfty$ and multiplying a number $<1$ by itself $n$ times (as $ntoinfty$) gets you to zero.

Notice that we are writing $xto0$, not $x=0$, as $x^1/x=0^infty$ is indeterminate.

share|cite|improve this answer

edited Jul 15 at 11:14

answered Jul 15 at 10:41

TheSimpliFire

9,69261951

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I hope that $x>0$ as you take your limit...?
  – Michael
  Jul 15 at 11:09
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Yes, as going from the negative side isn't defined.
  – TheSimpliFire
  Jul 15 at 11:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Perhaps it is better to write $lim_xrightarrow 0^+$.
  – Michael
  Jul 15 at 11:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Michael Edited.
  – TheSimpliFire
  Jul 15 at 11:14
  
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

up vote
5
down vote



accepted

We have $$lim_xto0^+x^1/x=0$$ since $x^1/x$ is not defined for all negative $x$.

This is because as $xto0$, $frac1xtoinfty$ and multiplying a number $<1$ by itself $n$ times (as $ntoinfty$) gets you to zero.

Notice that we are writing $xto0$, not $x=0$, as $x^1/x=0^infty$ is indeterminate.

share|cite|improve this answer

edited Jul 15 at 11:14

answered Jul 15 at 10:41

TheSimpliFire

9,69261951

We have $$lim_xto0^+x^1/x=0$$ since $x^1/x$ is not defined for all negative $x$.

This is because as $xto0$, $frac1xtoinfty$ and multiplying a number $<1$ by itself $n$ times (as $ntoinfty$) gets you to zero.

Notice that we are writing $xto0$, not $x=0$, as $x^1/x=0^infty$ is indeterminate.

share|cite|improve this answer

edited Jul 15 at 11:14

answered Jul 15 at 10:41

TheSimpliFire

9,69261951

share|cite|improve this answer

share|cite|improve this answer

edited Jul 15 at 11:14

edited Jul 15 at 11:14

edited Jul 15 at 11:14

answered Jul 15 at 10:41

TheSimpliFire

9,69261951

answered Jul 15 at 10:41

TheSimpliFire

9,69261951

answered Jul 15 at 10:41

TheSimpliFire

9,69261951

9,69261951

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I hope that $x>0$ as you take your limit...?
  – Michael
  Jul 15 at 11:09
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Yes, as going from the negative side isn't defined.
  – TheSimpliFire
  Jul 15 at 11:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Perhaps it is better to write $lim_xrightarrow 0^+$.
  – Michael
  Jul 15 at 11:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Michael Edited.
  – TheSimpliFire
  Jul 15 at 11:14
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I hope that $x>0$ as you take your limit...?
  – Michael
  Jul 15 at 11:09
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Yes, as going from the negative side isn't defined.
  – TheSimpliFire
  Jul 15 at 11:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Perhaps it is better to write $lim_xrightarrow 0^+$.
  – Michael
  Jul 15 at 11:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Michael Edited.
  – TheSimpliFire
  Jul 15 at 11:14
  
  

  
  
  
  

1

1

I hope that $x>0$ as you take your limit...?
– Michael
Jul 15 at 11:09

I hope that $x>0$ as you take your limit...?
– Michael
Jul 15 at 11:09

3

3

Yes, as going from the negative side isn't defined.
– TheSimpliFire
Jul 15 at 11:09

Yes, as going from the negative side isn't defined.
– TheSimpliFire
Jul 15 at 11:09

1

1

Perhaps it is better to write $lim_xrightarrow 0^+$.
– Michael
Jul 15 at 11:13

Perhaps it is better to write $lim_xrightarrow 0^+$.
– Michael
Jul 15 at 11:13

1

1

@Michael Edited.
– TheSimpliFire
Jul 15 at 11:14

@Michael Edited.
– TheSimpliFire
Jul 15 at 11:14

add a comment | 

up vote
2
down vote

The limit dose not exist because for negative values of $x$, $x^1/x$ is not defined.

share|cite|improve this answer

edited Jul 15 at 11:08

answered Jul 15 at 10:55

Mohammad Riazi-Kermani

27.6k41852

add a comment | 

up vote
2
down vote

The limit dose not exist because for negative values of $x$, $x^1/x$ is not defined.

share|cite|improve this answer

edited Jul 15 at 11:08

answered Jul 15 at 10:55

Mohammad Riazi-Kermani

27.6k41852

add a comment | 

up vote
2
down vote

up vote
2
down vote

The limit dose not exist because for negative values of $x$, $x^1/x$ is not defined.

share|cite|improve this answer

edited Jul 15 at 11:08

answered Jul 15 at 10:55

Mohammad Riazi-Kermani

27.6k41852

The limit dose not exist because for negative values of $x$, $x^1/x$ is not defined.

share|cite|improve this answer

edited Jul 15 at 11:08

answered Jul 15 at 10:55

Mohammad Riazi-Kermani

27.6k41852

share|cite|improve this answer

share|cite|improve this answer

edited Jul 15 at 11:08

edited Jul 15 at 11:08

edited Jul 15 at 11:08

answered Jul 15 at 10:55

Mohammad Riazi-Kermani

27.6k41852

answered Jul 15 at 10:55

Mohammad Riazi-Kermani

27.6k41852

answered Jul 15 at 10:55

Mohammad Riazi-Kermani

27.6k41852

27.6k41852

add a comment | 

add a comment | 

up vote
1
down vote

Let $L = lim_x to 0rootxof x$. Then, taking logarithm on both sides,

$$beginaligned
ln L = lim_x to 0fraclnxx
endaligned$$

This quantity is not defined, that is, it is an indeterminate of the form $fracinfty0$.

share|cite|improve this answer

edited Jul 15 at 11:22

answered Jul 15 at 10:46

Quasar

697412

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Have you tried L'Hopital?
  – TheSimpliFire
  Jul 15 at 11:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  As far as I remember, L' Hopital is applicable only when the $lim_x to cf(x) = lim_x to c g(x) =0$ or $pm infty$. So, it wouldn't be applicable in this case.
  – Quasar
  Jul 15 at 11:19
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

Let $L = lim_x to 0rootxof x$. Then, taking logarithm on both sides,

$$beginaligned
ln L = lim_x to 0fraclnxx
endaligned$$

This quantity is not defined, that is, it is an indeterminate of the form $fracinfty0$.

share|cite|improve this answer

edited Jul 15 at 11:22

answered Jul 15 at 10:46

Quasar

697412

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Have you tried L'Hopital?
  – TheSimpliFire
  Jul 15 at 11:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  As far as I remember, L' Hopital is applicable only when the $lim_x to cf(x) = lim_x to c g(x) =0$ or $pm infty$. So, it wouldn't be applicable in this case.
  – Quasar
  Jul 15 at 11:19
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Let $L = lim_x to 0rootxof x$. Then, taking logarithm on both sides,

$$beginaligned
ln L = lim_x to 0fraclnxx
endaligned$$

This quantity is not defined, that is, it is an indeterminate of the form $fracinfty0$.

share|cite|improve this answer

edited Jul 15 at 11:22

answered Jul 15 at 10:46

Quasar

697412

Let $L = lim_x to 0rootxof x$. Then, taking logarithm on both sides,

$$beginaligned
ln L = lim_x to 0fraclnxx
endaligned$$

This quantity is not defined, that is, it is an indeterminate of the form $fracinfty0$.

share|cite|improve this answer

edited Jul 15 at 11:22

answered Jul 15 at 10:46

Quasar

697412

share|cite|improve this answer

share|cite|improve this answer

edited Jul 15 at 11:22

edited Jul 15 at 11:22

edited Jul 15 at 11:22

answered Jul 15 at 10:46

Quasar

697412

answered Jul 15 at 10:46

Quasar

697412

answered Jul 15 at 10:46

Quasar

697412

697412

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Have you tried L'Hopital?
  – TheSimpliFire
  Jul 15 at 11:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  As far as I remember, L' Hopital is applicable only when the $lim_x to cf(x) = lim_x to c g(x) =0$ or $pm infty$. So, it wouldn't be applicable in this case.
  – Quasar
  Jul 15 at 11:19
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Have you tried L'Hopital?
  – TheSimpliFire
  Jul 15 at 11:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  As far as I remember, L' Hopital is applicable only when the $lim_x to cf(x) = lim_x to c g(x) =0$ or $pm infty$. So, it wouldn't be applicable in this case.
  – Quasar
  Jul 15 at 11:19
  
  

  
  
  
  

2

2

Have you tried L'Hopital?
– TheSimpliFire
Jul 15 at 11:05

Have you tried L'Hopital?
– TheSimpliFire
Jul 15 at 11:05

1

1

As far as I remember, L' Hopital is applicable only when the $lim_x to cf(x) = lim_x to c g(x) =0$ or $pm infty$. So, it wouldn't be applicable in this case.
– Quasar
Jul 15 at 11:19

As far as I remember, L' Hopital is applicable only when the $lim_x to cf(x) = lim_x to c g(x) =0$ or $pm infty$. So, it wouldn't be applicable in this case.
– Quasar
Jul 15 at 11:19

add a comment | 

up vote
1
down vote

No. It is undefined, and here's why.

^{You write $sqrt[0]n$, but I will write it as $sqrt[0]x$ with respect to the limit (because it has a variable $x$).}

Regarding the zeroth root,$$beginaligntextLet ;;,y&=sqrt[0]x, \ textthen ;y^0&=x.endalign$$

Since anything raised to the power of $0$ is $1$, it follows then that $x=1$. $$therefore sqrt[0]xtext is undefinedLeftrightarrow xneq 1.tag$therefore sqrt[0]0$ is undefined.$$ But $y$ could be any number, too, so this means $sqrt[0]x$ has no definite value, even if $xneq 1$.

$$therefore sqrt[0]xtext is undefined forall x.$$

---

Regarding the limit,
$$Lambda=lim_xto0x^frac 1x$$ this is of course equal to $0$. $overbracetextSince $0^n=0$ for all $nneq 0$,^large Statement (1)$ then as $xto 0$, it follows that $x^frac 1xto 0$, and by definition of a limit, we get that $Lambda = 0$.

But this does not mean that $1div 0$ is defined — it is not; division by zero is undefined, regardless of the value of the numerator. We just have to consider the limit, $$beginalignLambda_0=lim_xto0frac 1x&=infty, \ because lim_xtoinftyfrac 1x&=0.endalign$$ Therefore, when evaluating the limit $Lambda$, we notice that $1div xto infty$.

And thus, by Statement $(1)$, it is fully clear that $Lambda = 0$; id est, $$lim_xto 0frac 1x=0.$$

---

If you are still unsure and/or confused, take a look at the best answer (in my opinion).

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edited Jul 15 at 11:53

answered Jul 15 at 11:36

user477343

4,21731039

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up vote
1
down vote

No. It is undefined, and here's why.

^{You write $sqrt[0]n$, but I will write it as $sqrt[0]x$ with respect to the limit (because it has a variable $x$).}

Regarding the zeroth root,$$beginaligntextLet ;;,y&=sqrt[0]x, \ textthen ;y^0&=x.endalign$$

Since anything raised to the power of $0$ is $1$, it follows then that $x=1$. $$therefore sqrt[0]xtext is undefinedLeftrightarrow xneq 1.tag$therefore sqrt[0]0$ is undefined.$$ But $y$ could be any number, too, so this means $sqrt[0]x$ has no definite value, even if $xneq 1$.

$$therefore sqrt[0]xtext is undefined forall x.$$

---

Regarding the limit,
$$Lambda=lim_xto0x^frac 1x$$ this is of course equal to $0$. $overbracetextSince $0^n=0$ for all $nneq 0$,^large Statement (1)$ then as $xto 0$, it follows that $x^frac 1xto 0$, and by definition of a limit, we get that $Lambda = 0$.

But this does not mean that $1div 0$ is defined — it is not; division by zero is undefined, regardless of the value of the numerator. We just have to consider the limit, $$beginalignLambda_0=lim_xto0frac 1x&=infty, \ because lim_xtoinftyfrac 1x&=0.endalign$$ Therefore, when evaluating the limit $Lambda$, we notice that $1div xto infty$.

And thus, by Statement $(1)$, it is fully clear that $Lambda = 0$; id est, $$lim_xto 0frac 1x=0.$$

---

If you are still unsure and/or confused, take a look at the best answer (in my opinion).

share|cite|improve this answer

edited Jul 15 at 11:53

answered Jul 15 at 11:36

user477343

4,21731039

add a comment | 

up vote
1
down vote

up vote
1
down vote

No. It is undefined, and here's why.

^{You write $sqrt[0]n$, but I will write it as $sqrt[0]x$ with respect to the limit (because it has a variable $x$).}

Regarding the zeroth root,$$beginaligntextLet ;;,y&=sqrt[0]x, \ textthen ;y^0&=x.endalign$$

Since anything raised to the power of $0$ is $1$, it follows then that $x=1$. $$therefore sqrt[0]xtext is undefinedLeftrightarrow xneq 1.tag$therefore sqrt[0]0$ is undefined.$$ But $y$ could be any number, too, so this means $sqrt[0]x$ has no definite value, even if $xneq 1$.

$$therefore sqrt[0]xtext is undefined forall x.$$

---

Regarding the limit,
$$Lambda=lim_xto0x^frac 1x$$ this is of course equal to $0$. $overbracetextSince $0^n=0$ for all $nneq 0$,^large Statement (1)$ then as $xto 0$, it follows that $x^frac 1xto 0$, and by definition of a limit, we get that $Lambda = 0$.

But this does not mean that $1div 0$ is defined — it is not; division by zero is undefined, regardless of the value of the numerator. We just have to consider the limit, $$beginalignLambda_0=lim_xto0frac 1x&=infty, \ because lim_xtoinftyfrac 1x&=0.endalign$$ Therefore, when evaluating the limit $Lambda$, we notice that $1div xto infty$.

And thus, by Statement $(1)$, it is fully clear that $Lambda = 0$; id est, $$lim_xto 0frac 1x=0.$$

---

If you are still unsure and/or confused, take a look at the best answer (in my opinion).

share|cite|improve this answer

edited Jul 15 at 11:53

answered Jul 15 at 11:36

user477343

4,21731039

No. It is undefined, and here's why.

^{You write $sqrt[0]n$, but I will write it as $sqrt[0]x$ with respect to the limit (because it has a variable $x$).}

Regarding the zeroth root,$$beginaligntextLet ;;,y&=sqrt[0]x, \ textthen ;y^0&=x.endalign$$

Since anything raised to the power of $0$ is $1$, it follows then that $x=1$. $$therefore sqrt[0]xtext is undefinedLeftrightarrow xneq 1.tag$therefore sqrt[0]0$ is undefined.$$ But $y$ could be any number, too, so this means $sqrt[0]x$ has no definite value, even if $xneq 1$.

$$therefore sqrt[0]xtext is undefined forall x.$$

---

Regarding the limit,
$$Lambda=lim_xto0x^frac 1x$$ this is of course equal to $0$. $overbracetextSince $0^n=0$ for all $nneq 0$,^large Statement (1)$ then as $xto 0$, it follows that $x^frac 1xto 0$, and by definition of a limit, we get that $Lambda = 0$.

But this does not mean that $1div 0$ is defined — it is not; division by zero is undefined, regardless of the value of the numerator. We just have to consider the limit, $$beginalignLambda_0=lim_xto0frac 1x&=infty, \ because lim_xtoinftyfrac 1x&=0.endalign$$ Therefore, when evaluating the limit $Lambda$, we notice that $1div xto infty$.

And thus, by Statement $(1)$, it is fully clear that $Lambda = 0$; id est, $$lim_xto 0frac 1x=0.$$

---

If you are still unsure and/or confused, take a look at the best answer (in my opinion).

share|cite|improve this answer

edited Jul 15 at 11:53

answered Jul 15 at 11:36

user477343

4,21731039

share|cite|improve this answer

share|cite|improve this answer

edited Jul 15 at 11:53

edited Jul 15 at 11:53

edited Jul 15 at 11:53

answered Jul 15 at 11:36

user477343

4,21731039

answered Jul 15 at 11:36

user477343

4,21731039

answered Jul 15 at 11:36

user477343

4,21731039

4,21731039

add a comment | 

add a comment |