Mixing
Is it appropriate to say that $lim_{x to 4}{(sqrt x-2)}$ is both $0$ and $-4$?
$begingroup$
Our teachers seem to ignore that, but if I take a limit of a function with square root, for example:
$$lim_{x to 4}{(sqrt x-2)}$$ is it appropriate to say that there are two answers? $0$ and $-4$?
Because in the solutions they gave us, nobody mentions the fact that $sqrt {x^2}=|x|$, and they only say that the answer is the positive result. In the example - $0$.
Is it correct? if so, why?
Thank you.
limits roots radicals
edited Jan 13 at 11:13
Blue
48.4k870154
asked Jan 13 at 10:39
NetanelNetanel
974
$endgroup$
|
show 3 more comments
$begingroup$
Our teachers seem to ignore that, but if I take a limit of a function with square root, for example:
$$lim_{x to 4}{(sqrt x-2)}$$ is it appropriate to say that there are two answers? $0$ and $-4$?
Because in the solutions they gave us, nobody mentions the fact that $sqrt {x^2}=|x|$, and they only say that the answer is the positive result. In the example - $0$.
Is it correct? if so, why?
Thank you.
limits roots radicals
edited Jan 13 at 11:13
Blue
48.4k870154
asked Jan 13 at 10:39
NetanelNetanel
974
$endgroup$
$begingroup$
$sqrt{x^2} = |x|$ is in the definition of square root.
$endgroup$
– OmG
Jan 13 at 10:41
$begingroup$
no a limit of real numbers can only have one value. Note that $sqrt{x}>0$ so there's no way you get a limit that is less than $-2$.
$endgroup$
– Yanko
Jan 13 at 10:44
4
$begingroup$
No, by convention/definition, the square root function always returns a non-negative value, so $sqrt{x} = vert xvert geq 0$. Here, $sqrt{4} = +2$ only.
$endgroup$
– KM101
Jan 13 at 10:48
$begingroup$
@KM101 I don't mean to be picky here, but I think you mean $sqrt{x^{color{red}{2}}}=|x|geq0$ for which you did not write the $2$, but we all know what you mean ;)
$endgroup$
– user477343
Jan 13 at 11:17
3
$begingroup$
It's true that $x^2=4$ has two solutions: $x=pm 2$. (More generally, $x^2=m$ has two solutions $x=pm sqrt{m}$. (Well, it's only one solution if $m=0$, but I digress...)) That is not the same thing as saying $sqrt{4}$ has two values, $pm 2$. When we write "$sqrt{4}$", we mean the "principal square root of $4$"; that is, the non-negative number whose square is $4$; that number is $2$ and only $2$. (Things get a little messier among the complex numbers, but that's a discussion for another time.)
$endgroup$
– Blue
Jan 13 at 11:21
|
show 3 more comments
$begingroup$
Our teachers seem to ignore that, but if I take a limit of a function with square root, for example:
$$lim_{x to 4}{(sqrt x-2)}$$ is it appropriate to say that there are two answers? $0$ and $-4$?
Because in the solutions they gave us, nobody mentions the fact that $sqrt {x^2}=|x|$, and they only say that the answer is the positive result. In the example - $0$.
Is it correct? if so, why?
Thank you.
limits roots radicals
edited Jan 13 at 11:13
Blue
48.4k870154
asked Jan 13 at 10:39
NetanelNetanel
974
$endgroup$
Our teachers seem to ignore that, but if I take a limit of a function with square root, for example:
$$lim_{x to 4}{(sqrt x-2)}$$ is it appropriate to say that there are two answers? $0$ and $-4$?
Because in the solutions they gave us, nobody mentions the fact that $sqrt {x^2}=|x|$, and they only say that the answer is the positive result. In the example - $0$.
Is it correct? if so, why?
Thank you.
limits roots radicals
limits roots radicals
edited Jan 13 at 11:13
Blue
48.4k870154
asked Jan 13 at 10:39
NetanelNetanel
974
edited Jan 13 at 11:13
Blue
48.4k870154
asked Jan 13 at 10:39
NetanelNetanel
974
edited Jan 13 at 11:13
Blue
48.4k870154
edited Jan 13 at 11:13
Blue
48.4k870154
edited Jan 13 at 11:13
Blue
48.4k870154
48.4k870154
asked Jan 13 at 10:39
NetanelNetanel
974
asked Jan 13 at 10:39
NetanelNetanel
974
asked Jan 13 at 10:39
NetanelNetanel
974
974
$begingroup$
$sqrt{x^2} = |x|$ is in the definition of square root.
$endgroup$
– OmG
Jan 13 at 10:41
$begingroup$
no a limit of real numbers can only have one value. Note that $sqrt{x}>0$ so there's no way you get a limit that is less than $-2$.
$endgroup$
– Yanko
Jan 13 at 10:44
4
$begingroup$
No, by convention/definition, the square root function always returns a non-negative value, so $sqrt{x} = vert xvert geq 0$. Here, $sqrt{4} = +2$ only.
$endgroup$
– KM101
Jan 13 at 10:48
$begingroup$
@KM101 I don't mean to be picky here, but I think you mean $sqrt{x^{color{red}{2}}}=|x|geq0$ for which you did not write the $2$, but we all know what you mean ;)
$endgroup$
– user477343
Jan 13 at 11:17
3
$begingroup$
It's true that $x^2=4$ has two solutions: $x=pm 2$. (More generally, $x^2=m$ has two solutions $x=pm sqrt{m}$. (Well, it's only one solution if $m=0$, but I digress...)) That is not the same thing as saying $sqrt{4}$ has two values, $pm 2$. When we write "$sqrt{4}$", we mean the "principal square root of $4$"; that is, the non-negative number whose square is $4$; that number is $2$ and only $2$. (Things get a little messier among the complex numbers, but that's a discussion for another time.)
$endgroup$
– Blue
Jan 13 at 11:21
|
show 3 more comments
$begingroup$
$sqrt{x^2} = |x|$ is in the definition of square root.
$endgroup$
– OmG
Jan 13 at 10:41
$begingroup$
no a limit of real numbers can only have one value. Note that $sqrt{x}>0$ so there's no way you get a limit that is less than $-2$.
$endgroup$
– Yanko
Jan 13 at 10:44
4
$begingroup$
No, by convention/definition, the square root function always returns a non-negative value, so $sqrt{x} = vert xvert geq 0$. Here, $sqrt{4} = +2$ only.
$endgroup$
– KM101
Jan 13 at 10:48
$begingroup$
@KM101 I don't mean to be picky here, but I think you mean $sqrt{x^{color{red}{2}}}=|x|geq0$ for which you did not write the $2$, but we all know what you mean ;)
$endgroup$
– user477343
Jan 13 at 11:17
3
$begingroup$
It's true that $x^2=4$ has two solutions: $x=pm 2$. (More generally, $x^2=m$ has two solutions $x=pm sqrt{m}$. (Well, it's only one solution if $m=0$, but I digress...)) That is not the same thing as saying $sqrt{4}$ has two values, $pm 2$. When we write "$sqrt{4}$", we mean the "principal square root of $4$"; that is, the non-negative number whose square is $4$; that number is $2$ and only $2$. (Things get a little messier among the complex numbers, but that's a discussion for another time.)
$endgroup$
– Blue
Jan 13 at 11:21
$begingroup$
$sqrt{x^2} = |x|$ is in the definition of square root.
$endgroup$
– OmG
Jan 13 at 10:41
$begingroup$
$sqrt{x^2} = |x|$ is in the definition of square root.
$endgroup$
– OmG
Jan 13 at 10:41
$begingroup$
no a limit of real numbers can only have one value. Note that $sqrt{x}>0$ so there's no way you get a limit that is less than $-2$.
$endgroup$
– Yanko
Jan 13 at 10:44
$begingroup$
no a limit of real numbers can only have one value. Note that $sqrt{x}>0$ so there's no way you get a limit that is less than $-2$.
$endgroup$
– Yanko
Jan 13 at 10:44
4
4
$begingroup$
No, by convention/definition, the square root function always returns a non-negative value, so $sqrt{x} = vert xvert geq 0$. Here, $sqrt{4} = +2$ only.
$endgroup$
– KM101
Jan 13 at 10:48
$begingroup$
No, by convention/definition, the square root function always returns a non-negative value, so $sqrt{x} = vert xvert geq 0$. Here, $sqrt{4} = +2$ only.
$endgroup$
– KM101
Jan 13 at 10:48
$begingroup$
@KM101 I don't mean to be picky here, but I think you mean $sqrt{x^{color{red}{2}}}=|x|geq0$ for which you did not write the $2$, but we all know what you mean ;)
$endgroup$
– user477343
Jan 13 at 11:17
$begingroup$
@KM101 I don't mean to be picky here, but I think you mean $sqrt{x^{color{red}{2}}}=|x|geq0$ for which you did not write the $2$, but we all know what you mean ;)
$endgroup$
– user477343
Jan 13 at 11:17
3
3
$begingroup$
It's true that $x^2=4$ has two solutions: $x=pm 2$. (More generally, $x^2=m$ has two solutions $x=pm sqrt{m}$. (Well, it's only one solution if $m=0$, but I digress...)) That is not the same thing as saying $sqrt{4}$ has two values, $pm 2$. When we write "$sqrt{4}$", we mean the "principal square root of $4$"; that is, the non-negative number whose square is $4$; that number is $2$ and only $2$. (Things get a little messier among the complex numbers, but that's a discussion for another time.)
$endgroup$
– Blue
Jan 13 at 11:21
$begingroup$
It's true that $x^2=4$ has two solutions: $x=pm 2$. (More generally, $x^2=m$ has two solutions $x=pm sqrt{m}$. (Well, it's only one solution if $m=0$, but I digress...)) That is not the same thing as saying $sqrt{4}$ has two values, $pm 2$. When we write "$sqrt{4}$", we mean the "principal square root of $4$"; that is, the non-negative number whose square is $4$; that number is $2$ and only $2$. (Things get a little messier among the complex numbers, but that's a discussion for another time.)
$endgroup$
– Blue
Jan 13 at 11:21
|
show 3 more comments
2 Answers
2
active
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$begingroup$
This is a common misconception about the function $sqrt{cdot},:[0,infty)toBbb R$ as the "inverse" of $f(x)=x^2$.
In general, a function can have an inverse only when it is injective but $f(x)=x^2$ is not injective (e.g. $f(2)=f(-2) = 4$ ), so strictly speaking its inverse doesn't exist.
However, we can restrict its domain to $[0,infty)$ and the function $flvert_{[0,infty)}$ would be injective there, hence we can define $$sqrt{x}:=flvert_{[0,infty)}^{-1}.$$ The function $sqrt{x}$ thus takes only one positive value for each input $xge 0$ (e.g. $sqrt{4}=2$) even though there are two numbers such that $x^2=4$.
Remark: There are some cases where $sqrt{cdot}$ is defined to be multivalued, e.g. in complex analysis. However, $sqrt{cdot}$ wouldn't be a "function" in that case. Note also that we could very well define
$$sqrt{x}:=flvert_{(-infty,0]}^{-1}$$
and we'd have $sqrt{4} = -2$ according to this definition. This is not popular for obvious reasons.
answered Jan 13 at 11:53
BigbearZzzBigbearZzz
8,72621652
$endgroup$
add a comment |
$begingroup$
According to definition of principal square root this is impossible. The principal square root of $xge0$ is denoted by $sqrt x$ and is always non-negative.
answered Jan 13 at 11:57
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
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2 Answers
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$begingroup$
This is a common misconception about the function $sqrt{cdot},:[0,infty)toBbb R$ as the "inverse" of $f(x)=x^2$.
In general, a function can have an inverse only when it is injective but $f(x)=x^2$ is not injective (e.g. $f(2)=f(-2) = 4$ ), so strictly speaking its inverse doesn't exist.
However, we can restrict its domain to $[0,infty)$ and the function $flvert_{[0,infty)}$ would be injective there, hence we can define $$sqrt{x}:=flvert_{[0,infty)}^{-1}.$$ The function $sqrt{x}$ thus takes only one positive value for each input $xge 0$ (e.g. $sqrt{4}=2$) even though there are two numbers such that $x^2=4$.
Remark: There are some cases where $sqrt{cdot}$ is defined to be multivalued, e.g. in complex analysis. However, $sqrt{cdot}$ wouldn't be a "function" in that case. Note also that we could very well define
$$sqrt{x}:=flvert_{(-infty,0]}^{-1}$$
and we'd have $sqrt{4} = -2$ according to this definition. This is not popular for obvious reasons.
answered Jan 13 at 11:53
BigbearZzzBigbearZzz
8,72621652
$endgroup$
add a comment |
$begingroup$
This is a common misconception about the function $sqrt{cdot},:[0,infty)toBbb R$ as the "inverse" of $f(x)=x^2$.
In general, a function can have an inverse only when it is injective but $f(x)=x^2$ is not injective (e.g. $f(2)=f(-2) = 4$ ), so strictly speaking its inverse doesn't exist.
However, we can restrict its domain to $[0,infty)$ and the function $flvert_{[0,infty)}$ would be injective there, hence we can define $$sqrt{x}:=flvert_{[0,infty)}^{-1}.$$ The function $sqrt{x}$ thus takes only one positive value for each input $xge 0$ (e.g. $sqrt{4}=2$) even though there are two numbers such that $x^2=4$.
Remark: There are some cases where $sqrt{cdot}$ is defined to be multivalued, e.g. in complex analysis. However, $sqrt{cdot}$ wouldn't be a "function" in that case. Note also that we could very well define
$$sqrt{x}:=flvert_{(-infty,0]}^{-1}$$
and we'd have $sqrt{4} = -2$ according to this definition. This is not popular for obvious reasons.
answered Jan 13 at 11:53
BigbearZzzBigbearZzz
8,72621652
$endgroup$
add a comment |
$begingroup$
This is a common misconception about the function $sqrt{cdot},:[0,infty)toBbb R$ as the "inverse" of $f(x)=x^2$.
In general, a function can have an inverse only when it is injective but $f(x)=x^2$ is not injective (e.g. $f(2)=f(-2) = 4$ ), so strictly speaking its inverse doesn't exist.
However, we can restrict its domain to $[0,infty)$ and the function $flvert_{[0,infty)}$ would be injective there, hence we can define $$sqrt{x}:=flvert_{[0,infty)}^{-1}.$$ The function $sqrt{x}$ thus takes only one positive value for each input $xge 0$ (e.g. $sqrt{4}=2$) even though there are two numbers such that $x^2=4$.
Remark: There are some cases where $sqrt{cdot}$ is defined to be multivalued, e.g. in complex analysis. However, $sqrt{cdot}$ wouldn't be a "function" in that case. Note also that we could very well define
$$sqrt{x}:=flvert_{(-infty,0]}^{-1}$$
and we'd have $sqrt{4} = -2$ according to this definition. This is not popular for obvious reasons.
answered Jan 13 at 11:53
BigbearZzzBigbearZzz
8,72621652
$endgroup$
This is a common misconception about the function $sqrt{cdot},:[0,infty)toBbb R$ as the "inverse" of $f(x)=x^2$.
In general, a function can have an inverse only when it is injective but $f(x)=x^2$ is not injective (e.g. $f(2)=f(-2) = 4$ ), so strictly speaking its inverse doesn't exist.
However, we can restrict its domain to $[0,infty)$ and the function $flvert_{[0,infty)}$ would be injective there, hence we can define $$sqrt{x}:=flvert_{[0,infty)}^{-1}.$$ The function $sqrt{x}$ thus takes only one positive value for each input $xge 0$ (e.g. $sqrt{4}=2$) even though there are two numbers such that $x^2=4$.
Remark: There are some cases where $sqrt{cdot}$ is defined to be multivalued, e.g. in complex analysis. However, $sqrt{cdot}$ wouldn't be a "function" in that case. Note also that we could very well define
$$sqrt{x}:=flvert_{(-infty,0]}^{-1}$$
and we'd have $sqrt{4} = -2$ according to this definition. This is not popular for obvious reasons.
answered Jan 13 at 11:53
BigbearZzzBigbearZzz
8,72621652
edited Jan 13 at 13:19
answered Jan 13 at 11:53
BigbearZzzBigbearZzz
8,72621652
answered Jan 13 at 11:53
BigbearZzzBigbearZzz
8,72621652
answered Jan 13 at 11:53
BigbearZzzBigbearZzz
8,72621652
8,72621652
add a comment |
add a comment |
$begingroup$
According to definition of principal square root this is impossible. The principal square root of $xge0$ is denoted by $sqrt x$ and is always non-negative.
answered Jan 13 at 11:57
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
According to definition of principal square root this is impossible. The principal square root of $xge0$ is denoted by $sqrt x$ and is always non-negative.
answered Jan 13 at 11:57
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
According to definition of principal square root this is impossible. The principal square root of $xge0$ is denoted by $sqrt x$ and is always non-negative.
answered Jan 13 at 11:57
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
According to definition of principal square root this is impossible. The principal square root of $xge0$ is denoted by $sqrt x$ and is always non-negative.
answered Jan 13 at 11:57
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 13 at 11:57
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 13 at 11:57
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 13 at 11:57
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
add a comment |
add a comment |
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$begingroup$
$sqrt{x^2} = |x|$ is in the definition of square root.
$endgroup$
– OmG
Jan 13 at 10:41
$begingroup$
no a limit of real numbers can only have one value. Note that $sqrt{x}>0$ so there's no way you get a limit that is less than $-2$.
$endgroup$
– Yanko
Jan 13 at 10:44
4
$begingroup$
No, by convention/definition, the square root function always returns a non-negative value, so $sqrt{x} = vert xvert geq 0$. Here, $sqrt{4} = +2$ only.
$endgroup$
– KM101
Jan 13 at 10:48
$begingroup$
@KM101 I don't mean to be picky here, but I think you mean $sqrt{x^{color{red}{2}}}=|x|geq0$ for which you did not write the $2$, but we all know what you mean ;)
$endgroup$
– user477343
Jan 13 at 11:17
3
$begingroup$
It's true that $x^2=4$ has two solutions: $x=pm 2$. (More generally, $x^2=m$ has two solutions $x=pm sqrt{m}$. (Well, it's only one solution if $m=0$, but I digress...)) That is not the same thing as saying $sqrt{4}$ has two values, $pm 2$. When we write "$sqrt{4}$", we mean the "principal square root of $4$"; that is, the non-negative number whose square is $4$; that number is $2$ and only $2$. (Things get a little messier among the complex numbers, but that's a discussion for another time.)
$endgroup$
– Blue
Jan 13 at 11:21