existence of limit when both RHL and LHL approachees Positive infinity
We know that if $$lim_{x to a^+}f(x)=lim_{x to a^-}f(x)=L$$ Then $$lim_{x to a}f(x) =L$$ if $L$ is finite
But if $$lim_{x to a^+}f(x) to +infty$$ and
$$lim_{x to a^-}f(x) to +infty$$
Can we say $$lim_{x to a} f(x)$$ Does not exists since we cannot compare two infinities.
calculus limits
asked Jan 5 '16 at 4:15
Umesh shankar
2,56331219
add a comment |
We know that if $$lim_{x to a^+}f(x)=lim_{x to a^-}f(x)=L$$ Then $$lim_{x to a}f(x) =L$$ if $L$ is finite
But if $$lim_{x to a^+}f(x) to +infty$$ and
$$lim_{x to a^-}f(x) to +infty$$
Can we say $$lim_{x to a} f(x)$$ Does not exists since we cannot compare two infinities.
calculus limits
asked Jan 5 '16 at 4:15
Umesh shankar
2,56331219
What do you mean "we cannot compare two infinities"? The statement $lim_{x to a} f(x) = infty$ is defined in terms of lower bounds of a function.
– T. Bongers
Jan 5 '16 at 4:17
what i mean is if $L$ is finite $L-L=0$, but if $L$ is $infty$, $infty-infty $ is indeterminate. Hence can we say Limit DNE?
– Umesh shankar
Jan 5 '16 at 4:19
1
No, the limit is equal to infinity. How would subtraction be involved?
– T. Bongers
Jan 5 '16 at 4:20
add a comment |
We know that if $$lim_{x to a^+}f(x)=lim_{x to a^-}f(x)=L$$ Then $$lim_{x to a}f(x) =L$$ if $L$ is finite
But if $$lim_{x to a^+}f(x) to +infty$$ and
$$lim_{x to a^-}f(x) to +infty$$
Can we say $$lim_{x to a} f(x)$$ Does not exists since we cannot compare two infinities.
calculus limits
asked Jan 5 '16 at 4:15
Umesh shankar
2,56331219
We know that if $$lim_{x to a^+}f(x)=lim_{x to a^-}f(x)=L$$ Then $$lim_{x to a}f(x) =L$$ if $L$ is finite
But if $$lim_{x to a^+}f(x) to +infty$$ and
$$lim_{x to a^-}f(x) to +infty$$
Can we say $$lim_{x to a} f(x)$$ Does not exists since we cannot compare two infinities.
calculus limits
calculus limits
asked Jan 5 '16 at 4:15
Umesh shankar
2,56331219
asked Jan 5 '16 at 4:15
Umesh shankar
2,56331219
asked Jan 5 '16 at 4:15
Umesh shankar
2,56331219
asked Jan 5 '16 at 4:15
Umesh shankar
2,56331219
asked Jan 5 '16 at 4:15
Umesh shankar
2,56331219
2,56331219
What do you mean "we cannot compare two infinities"? The statement $lim_{x to a} f(x) = infty$ is defined in terms of lower bounds of a function.
– T. Bongers
Jan 5 '16 at 4:17
what i mean is if $L$ is finite $L-L=0$, but if $L$ is $infty$, $infty-infty $ is indeterminate. Hence can we say Limit DNE?
– Umesh shankar
Jan 5 '16 at 4:19
1
No, the limit is equal to infinity. How would subtraction be involved?
– T. Bongers
Jan 5 '16 at 4:20
add a comment |
What do you mean "we cannot compare two infinities"? The statement $lim_{x to a} f(x) = infty$ is defined in terms of lower bounds of a function.
– T. Bongers
Jan 5 '16 at 4:17
what i mean is if $L$ is finite $L-L=0$, but if $L$ is $infty$, $infty-infty $ is indeterminate. Hence can we say Limit DNE?
– Umesh shankar
Jan 5 '16 at 4:19
1
No, the limit is equal to infinity. How would subtraction be involved?
– T. Bongers
Jan 5 '16 at 4:20
What do you mean "we cannot compare two infinities"? The statement $lim_{x to a} f(x) = infty$ is defined in terms of lower bounds of a function.
– T. Bongers
Jan 5 '16 at 4:17
What do you mean "we cannot compare two infinities"? The statement $lim_{x to a} f(x) = infty$ is defined in terms of lower bounds of a function.
– T. Bongers
Jan 5 '16 at 4:17
what i mean is if $L$ is finite $L-L=0$, but if $L$ is $infty$, $infty-infty $ is indeterminate. Hence can we say Limit DNE?
– Umesh shankar
Jan 5 '16 at 4:19
what i mean is if $L$ is finite $L-L=0$, but if $L$ is $infty$, $infty-infty $ is indeterminate. Hence can we say Limit DNE?
– Umesh shankar
Jan 5 '16 at 4:19
1
1
No, the limit is equal to infinity. How would subtraction be involved?
– T. Bongers
Jan 5 '16 at 4:20
No, the limit is equal to infinity. How would subtraction be involved?
– T. Bongers
Jan 5 '16 at 4:20
add a comment |
3 Answers
3
active
oldest
votes
Having $$lim_{x to a^+} f(x)=infty$$ alone is enough to conclude that the limit does not exist, since otherwise there is a finite $L$ such that $$lim_{x to a} f(x)=L,$$ which implies that $$lim_{x to a^+} f(x)=L.$$
answered Jan 5 '16 at 4:36
zwisch
12
Limits of $pm infty$ can exist.
– zhw.
May 14 '17 at 21:39
add a comment |
In general, we only say a limit 'exists' when it is finite. When we say $lim_{xto a}f(x)=+infty$ we mean that
$$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta),, f(x)>M tag{1}$$
In other words, $f$ becomes arbitrarily large near $a$ — all of $f$. Contrast this with '$f$ attains arbitrarily large value near $a$'.
When we say $lim_{xto a^+}f(x)=+infty$ we mean that
$$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x>a),, f(x)>Mtag{2}$$
In other words, $f$ becomes arbitrarily large in a right-neighborhood of $a$ — all of $f$, once again.
When we say $lim_{xto a^-}f(x)=+infty$ we mean that
$$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x<a),, f(x)>Mtag{3}$$
In other words, $f$ becomes arbitrarily large in a left-neighborhood of $a$ — all of $f$, once again.
Can you see how statements $(2)$ and $(3)$, combined, imply statement $(1)$?
answered Nov 6 '17 at 14:18
Fimpellizieri
17.2k11836
add a comment |
Can we say $lim limits_{xrightarrow a}f(x)$ does not exists since we cannot compare two infinities?
Yes, we can say that it does not exist at a point $a$ because plus and minus infinities are different behaviors (function growing without bound/decreasing without bound). Moreover, limits of functions that end up in plus or minus infinity actually do not exist either. However, we do write that a limit is equal to $pminfty$, but that's only done for the purposes of describing the way a function behaves. It's just supposed to be more informative that way, meaning it gives us more information about a function's behavior near a particular point.
answered Nov 6 '17 at 14:34
Mike R.
1,339212
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Having $$lim_{x to a^+} f(x)=infty$$ alone is enough to conclude that the limit does not exist, since otherwise there is a finite $L$ such that $$lim_{x to a} f(x)=L,$$ which implies that $$lim_{x to a^+} f(x)=L.$$
answered Jan 5 '16 at 4:36
zwisch
12
Limits of $pm infty$ can exist.
– zhw.
May 14 '17 at 21:39
add a comment |
Having $$lim_{x to a^+} f(x)=infty$$ alone is enough to conclude that the limit does not exist, since otherwise there is a finite $L$ such that $$lim_{x to a} f(x)=L,$$ which implies that $$lim_{x to a^+} f(x)=L.$$
answered Jan 5 '16 at 4:36
zwisch
12
Limits of $pm infty$ can exist.
– zhw.
May 14 '17 at 21:39
add a comment |
Having $$lim_{x to a^+} f(x)=infty$$ alone is enough to conclude that the limit does not exist, since otherwise there is a finite $L$ such that $$lim_{x to a} f(x)=L,$$ which implies that $$lim_{x to a^+} f(x)=L.$$
answered Jan 5 '16 at 4:36
zwisch
12
Having $$lim_{x to a^+} f(x)=infty$$ alone is enough to conclude that the limit does not exist, since otherwise there is a finite $L$ such that $$lim_{x to a} f(x)=L,$$ which implies that $$lim_{x to a^+} f(x)=L.$$
answered Jan 5 '16 at 4:36
zwisch
12
answered Jan 5 '16 at 4:36
zwisch
12
answered Jan 5 '16 at 4:36
zwisch
12
answered Jan 5 '16 at 4:36
zwisch
12
12
Limits of $pm infty$ can exist.
– zhw.
May 14 '17 at 21:39
add a comment |
Limits of $pm infty$ can exist.
– zhw.
May 14 '17 at 21:39
Limits of $pm infty$ can exist.
– zhw.
May 14 '17 at 21:39
Limits of $pm infty$ can exist.
– zhw.
May 14 '17 at 21:39
add a comment |
In general, we only say a limit 'exists' when it is finite. When we say $lim_{xto a}f(x)=+infty$ we mean that
$$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta),, f(x)>M tag{1}$$
In other words, $f$ becomes arbitrarily large near $a$ — all of $f$. Contrast this with '$f$ attains arbitrarily large value near $a$'.
When we say $lim_{xto a^+}f(x)=+infty$ we mean that
$$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x>a),, f(x)>Mtag{2}$$
In other words, $f$ becomes arbitrarily large in a right-neighborhood of $a$ — all of $f$, once again.
When we say $lim_{xto a^-}f(x)=+infty$ we mean that
$$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x<a),, f(x)>Mtag{3}$$
In other words, $f$ becomes arbitrarily large in a left-neighborhood of $a$ — all of $f$, once again.
Can you see how statements $(2)$ and $(3)$, combined, imply statement $(1)$?
answered Nov 6 '17 at 14:18
Fimpellizieri
17.2k11836
add a comment |
In general, we only say a limit 'exists' when it is finite. When we say $lim_{xto a}f(x)=+infty$ we mean that
$$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta),, f(x)>M tag{1}$$
In other words, $f$ becomes arbitrarily large near $a$ — all of $f$. Contrast this with '$f$ attains arbitrarily large value near $a$'.
When we say $lim_{xto a^+}f(x)=+infty$ we mean that
$$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x>a),, f(x)>Mtag{2}$$
In other words, $f$ becomes arbitrarily large in a right-neighborhood of $a$ — all of $f$, once again.
When we say $lim_{xto a^-}f(x)=+infty$ we mean that
$$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x<a),, f(x)>Mtag{3}$$
In other words, $f$ becomes arbitrarily large in a left-neighborhood of $a$ — all of $f$, once again.
Can you see how statements $(2)$ and $(3)$, combined, imply statement $(1)$?
answered Nov 6 '17 at 14:18
Fimpellizieri
17.2k11836
add a comment |
In general, we only say a limit 'exists' when it is finite. When we say $lim_{xto a}f(x)=+infty$ we mean that
$$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta),, f(x)>M tag{1}$$
In other words, $f$ becomes arbitrarily large near $a$ — all of $f$. Contrast this with '$f$ attains arbitrarily large value near $a$'.
When we say $lim_{xto a^+}f(x)=+infty$ we mean that
$$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x>a),, f(x)>Mtag{2}$$
In other words, $f$ becomes arbitrarily large in a right-neighborhood of $a$ — all of $f$, once again.
When we say $lim_{xto a^-}f(x)=+infty$ we mean that
$$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x<a),, f(x)>Mtag{3}$$
In other words, $f$ becomes arbitrarily large in a left-neighborhood of $a$ — all of $f$, once again.
Can you see how statements $(2)$ and $(3)$, combined, imply statement $(1)$?
answered Nov 6 '17 at 14:18
Fimpellizieri
17.2k11836
In general, we only say a limit 'exists' when it is finite. When we say $lim_{xto a}f(x)=+infty$ we mean that
$$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta),, f(x)>M tag{1}$$
In other words, $f$ becomes arbitrarily large near $a$ — all of $f$. Contrast this with '$f$ attains arbitrarily large value near $a$'.
When we say $lim_{xto a^+}f(x)=+infty$ we mean that
$$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x>a),, f(x)>Mtag{2}$$
In other words, $f$ becomes arbitrarily large in a right-neighborhood of $a$ — all of $f$, once again.
When we say $lim_{xto a^-}f(x)=+infty$ we mean that
$$(forall M>0),(exists delta > 0),(forall x ,text{ with }, 0<|x-a|<delta,text{ and }, x<a),, f(x)>Mtag{3}$$
In other words, $f$ becomes arbitrarily large in a left-neighborhood of $a$ — all of $f$, once again.
Can you see how statements $(2)$ and $(3)$, combined, imply statement $(1)$?
answered Nov 6 '17 at 14:18
Fimpellizieri
17.2k11836
answered Nov 6 '17 at 14:18
Fimpellizieri
17.2k11836
answered Nov 6 '17 at 14:18
Fimpellizieri
17.2k11836
answered Nov 6 '17 at 14:18
Fimpellizieri
17.2k11836
17.2k11836
add a comment |
add a comment |
Can we say $lim limits_{xrightarrow a}f(x)$ does not exists since we cannot compare two infinities?
Yes, we can say that it does not exist at a point $a$ because plus and minus infinities are different behaviors (function growing without bound/decreasing without bound). Moreover, limits of functions that end up in plus or minus infinity actually do not exist either. However, we do write that a limit is equal to $pminfty$, but that's only done for the purposes of describing the way a function behaves. It's just supposed to be more informative that way, meaning it gives us more information about a function's behavior near a particular point.
answered Nov 6 '17 at 14:34
Mike R.
1,339212
add a comment |
Can we say $lim limits_{xrightarrow a}f(x)$ does not exists since we cannot compare two infinities?
Yes, we can say that it does not exist at a point $a$ because plus and minus infinities are different behaviors (function growing without bound/decreasing without bound). Moreover, limits of functions that end up in plus or minus infinity actually do not exist either. However, we do write that a limit is equal to $pminfty$, but that's only done for the purposes of describing the way a function behaves. It's just supposed to be more informative that way, meaning it gives us more information about a function's behavior near a particular point.
answered Nov 6 '17 at 14:34
Mike R.
1,339212
add a comment |
Can we say $lim limits_{xrightarrow a}f(x)$ does not exists since we cannot compare two infinities?
Yes, we can say that it does not exist at a point $a$ because plus and minus infinities are different behaviors (function growing without bound/decreasing without bound). Moreover, limits of functions that end up in plus or minus infinity actually do not exist either. However, we do write that a limit is equal to $pminfty$, but that's only done for the purposes of describing the way a function behaves. It's just supposed to be more informative that way, meaning it gives us more information about a function's behavior near a particular point.
answered Nov 6 '17 at 14:34
Mike R.
1,339212
Can we say $lim limits_{xrightarrow a}f(x)$ does not exists since we cannot compare two infinities?
Yes, we can say that it does not exist at a point $a$ because plus and minus infinities are different behaviors (function growing without bound/decreasing without bound). Moreover, limits of functions that end up in plus or minus infinity actually do not exist either. However, we do write that a limit is equal to $pminfty$, but that's only done for the purposes of describing the way a function behaves. It's just supposed to be more informative that way, meaning it gives us more information about a function's behavior near a particular point.
answered Nov 6 '17 at 14:34
Mike R.
1,339212
edited Nov 6 '17 at 14:40
answered Nov 6 '17 at 14:34
Mike R.
1,339212
answered Nov 6 '17 at 14:34
Mike R.
1,339212
answered Nov 6 '17 at 14:34
Mike R.
1,339212
1,339212
add a comment |
add a comment |
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What do you mean "we cannot compare two infinities"? The statement $lim_{x to a} f(x) = infty$ is defined in terms of lower bounds of a function.
– T. Bongers
Jan 5 '16 at 4:17
what i mean is if $L$ is finite $L-L=0$, but if $L$ is $infty$, $infty-infty $ is indeterminate. Hence can we say Limit DNE?
– Umesh shankar
Jan 5 '16 at 4:19
1
No, the limit is equal to infinity. How would subtraction be involved?
– T. Bongers
Jan 5 '16 at 4:20