Mixing
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Why do the properties of $limlimits_{nrightarrowinfty}(x_n)$ still hold for $limlimits_{xrightarrow x_0}f(x)$
$begingroup$
Where is the transition of sequences and functions?
With properties I mean for example rules like the sandwich theorem
$$lim_{nrightarrowinfty}a_n=K, lim_{nrightarrowinfty}c_n=Ktext{ and }a_nleq b_nleq c_nforall_{ninmathbb{N}}Rightarrowlim_{nrightarrowinfty}b_n=K$$
Or that one can "pull out" the constant, i.e
$$lim_{nrightarrowinfty}a_n=KRightarrow lim_{nrightarrowinfty}ca_n=clim_{nrightarrowinfty}a_n=cK$$
To name a few.
sequences-and-series limits
edited Feb 10 at 23:36
Paul Frost
11.6k3934
asked Jan 23 at 12:50
RM777RM777
38312
$endgroup$
add a comment |
$begingroup$
Where is the transition of sequences and functions?
With properties I mean for example rules like the sandwich theorem
$$lim_{nrightarrowinfty}a_n=K, lim_{nrightarrowinfty}c_n=Ktext{ and }a_nleq b_nleq c_nforall_{ninmathbb{N}}Rightarrowlim_{nrightarrowinfty}b_n=K$$
Or that one can "pull out" the constant, i.e
$$lim_{nrightarrowinfty}a_n=KRightarrow lim_{nrightarrowinfty}ca_n=clim_{nrightarrowinfty}a_n=cK$$
To name a few.
sequences-and-series limits
edited Feb 10 at 23:36
Paul Frost
11.6k3934
asked Jan 23 at 12:50
RM777RM777
38312
$endgroup$
1
$begingroup$
This is intuitively clear if you think about limits in terms of arbitrarily close approximations. If you look at the parallel proofs using the definitions of each kind of limit you will see the same essential arguments.
$endgroup$
– Ethan Bolker
Jan 23 at 12:53
$begingroup$
If you are expecting a proof then, have a look at this: en.wikipedia.org/wiki/Squeeze_theorem. For the second proof just use the $epsilon-N$ definition of limits.
$endgroup$
– Sujit Bhattacharyya
Jan 23 at 12:58
$begingroup$
The definition of limit of sequences and limit of functions are very similar and hence one expects similar results. To be more formal, if you understand the proof for these results on sequences you should not have any trouble writing a proof for corresponding result on functions.
$endgroup$
– Paramanand Singh
Jan 24 at 3:28
add a comment |
$begingroup$
Where is the transition of sequences and functions?
With properties I mean for example rules like the sandwich theorem
$$lim_{nrightarrowinfty}a_n=K, lim_{nrightarrowinfty}c_n=Ktext{ and }a_nleq b_nleq c_nforall_{ninmathbb{N}}Rightarrowlim_{nrightarrowinfty}b_n=K$$
Or that one can "pull out" the constant, i.e
$$lim_{nrightarrowinfty}a_n=KRightarrow lim_{nrightarrowinfty}ca_n=clim_{nrightarrowinfty}a_n=cK$$
To name a few.
sequences-and-series limits
edited Feb 10 at 23:36
Paul Frost
11.6k3934
asked Jan 23 at 12:50
RM777RM777
38312
$endgroup$
Where is the transition of sequences and functions?
With properties I mean for example rules like the sandwich theorem
$$lim_{nrightarrowinfty}a_n=K, lim_{nrightarrowinfty}c_n=Ktext{ and }a_nleq b_nleq c_nforall_{ninmathbb{N}}Rightarrowlim_{nrightarrowinfty}b_n=K$$
Or that one can "pull out" the constant, i.e
$$lim_{nrightarrowinfty}a_n=KRightarrow lim_{nrightarrowinfty}ca_n=clim_{nrightarrowinfty}a_n=cK$$
To name a few.
sequences-and-series limits
sequences-and-series limits
edited Feb 10 at 23:36
Paul Frost
11.6k3934
asked Jan 23 at 12:50
RM777RM777
38312
edited Feb 10 at 23:36
Paul Frost
11.6k3934
asked Jan 23 at 12:50
RM777RM777
38312
edited Feb 10 at 23:36
Paul Frost
11.6k3934
edited Feb 10 at 23:36
Paul Frost
11.6k3934
edited Feb 10 at 23:36
Paul Frost
11.6k3934
11.6k3934
asked Jan 23 at 12:50
RM777RM777
38312
asked Jan 23 at 12:50
RM777RM777
38312
asked Jan 23 at 12:50
RM777RM777
38312
38312
1
$begingroup$
This is intuitively clear if you think about limits in terms of arbitrarily close approximations. If you look at the parallel proofs using the definitions of each kind of limit you will see the same essential arguments.
$endgroup$
– Ethan Bolker
Jan 23 at 12:53
$begingroup$
If you are expecting a proof then, have a look at this: en.wikipedia.org/wiki/Squeeze_theorem. For the second proof just use the $epsilon-N$ definition of limits.
$endgroup$
– Sujit Bhattacharyya
Jan 23 at 12:58
$begingroup$
The definition of limit of sequences and limit of functions are very similar and hence one expects similar results. To be more formal, if you understand the proof for these results on sequences you should not have any trouble writing a proof for corresponding result on functions.
$endgroup$
– Paramanand Singh
Jan 24 at 3:28
add a comment |
1
$begingroup$
This is intuitively clear if you think about limits in terms of arbitrarily close approximations. If you look at the parallel proofs using the definitions of each kind of limit you will see the same essential arguments.
$endgroup$
– Ethan Bolker
Jan 23 at 12:53
$begingroup$
If you are expecting a proof then, have a look at this: en.wikipedia.org/wiki/Squeeze_theorem. For the second proof just use the $epsilon-N$ definition of limits.
$endgroup$
– Sujit Bhattacharyya
Jan 23 at 12:58
$begingroup$
The definition of limit of sequences and limit of functions are very similar and hence one expects similar results. To be more formal, if you understand the proof for these results on sequences you should not have any trouble writing a proof for corresponding result on functions.
$endgroup$
– Paramanand Singh
Jan 24 at 3:28
1
1
$begingroup$
This is intuitively clear if you think about limits in terms of arbitrarily close approximations. If you look at the parallel proofs using the definitions of each kind of limit you will see the same essential arguments.
$endgroup$
– Ethan Bolker
Jan 23 at 12:53
$begingroup$
This is intuitively clear if you think about limits in terms of arbitrarily close approximations. If you look at the parallel proofs using the definitions of each kind of limit you will see the same essential arguments.
$endgroup$
– Ethan Bolker
Jan 23 at 12:53
$begingroup$
If you are expecting a proof then, have a look at this: en.wikipedia.org/wiki/Squeeze_theorem. For the second proof just use the $epsilon-N$ definition of limits.
$endgroup$
– Sujit Bhattacharyya
Jan 23 at 12:58
$begingroup$
If you are expecting a proof then, have a look at this: en.wikipedia.org/wiki/Squeeze_theorem. For the second proof just use the $epsilon-N$ definition of limits.
$endgroup$
– Sujit Bhattacharyya
Jan 23 at 12:58
$begingroup$
The definition of limit of sequences and limit of functions are very similar and hence one expects similar results. To be more formal, if you understand the proof for these results on sequences you should not have any trouble writing a proof for corresponding result on functions.
$endgroup$
– Paramanand Singh
Jan 24 at 3:28
$begingroup$
The definition of limit of sequences and limit of functions are very similar and hence one expects similar results. To be more formal, if you understand the proof for these results on sequences you should not have any trouble writing a proof for corresponding result on functions.
$endgroup$
– Paramanand Singh
Jan 24 at 3:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We go into some generality, when we approach what I would call the link between sequence and function limits.
Sequence limits are given by the $epsilon-N$ definition and function limits by the $epsilon-delta$ definition, if you recall.
Well, first, on the real line, let us prove the following proposition (I am calling this SFR, but you can give it any name you like : it is just nice):
SFR (Sequence-Function Relation) : Let $f : mathbb R to mathbb R$ and $x_0,L in mathbb R$. The following are equivalent :
$(1)displaystylelim_{x to x_0} f(x) = L$.
$(2)$ For every sequence ${x_n} subset mathbb R$ such that $x_n$ converges to $x_0$, we have that the sequence ${f(x_n)}$ converges to $L$.
Notation : $a_n to b$ means "the sequence $a_n$ converges to $b$".
Proof : Suppose $lim_{x to x_0} f(x) = L$. Fix a sequence $y_n to x_0$. We want to show that $f(y_n) to L$. For this , we start with $epsilon > 0$.
For that $epsilon$, we get a $delta > 0$ from $(1)$ such that whenever $|x_0-z| < delta$ we have $|L-f(z)| < epsilon$.
Now, for this $delta > 0$ we get from the fact that $y_n to x_0$, an $N in mathbb N$ such that $m > N implies |y_n -x_0| < delta$.
Therefore :
$$
m > N implies |y_n - x_0| < delta implies |f(y_n) - f(x_0)| < epsilon
$$
and hence $f(y_n) to f(x_0)$.
For the other way, we assume that $(1)$ is not true. So $lim_{x to x_0} f(x) neq L$. We now have to find a sequence $y_n to x_0$ such that $f(y_n) not to L$.
For this, we "negate" the definition of the limit existing : since the limit says that for all $epsilon > 0$ something happens, if the limit does not exist then for some $epsilon_0 > 0$ that something does not happen.
What does not happen? "I can find a $delta > 0$ so that ... happens" does not happen. In other words, you can't find a $delta > 0$ for this $epsilon_0$. Or, every $delta > 0$ fails for this $epsilon_0$.
And what does $delta$ being a failure mean? It means that for some point $x$ such that $|x-x_0| < delta$, we have $|f(x) - L|$ is not less than $epsilon_0$, or in other words $|f(x) - L| geq epsilon_0$.
So, $lim_{x to x_0} f(x) neq L$ means exactly this :
there exists $epsilon_0 > 0$ such that for all $delta > 0$, there exists $x$ such that $|x - x_0| < delta$ but $|f(x) - L| geq epsilon$.
Geometrically, this means : there is an $epsilon_0 >0$ such that you can find points as close to $x_0$ as you want, with the function value at that point at least $epsilon_0$ away from $L$.
With these tools, in the boxed quotes we take $delta = frac 1n$. For each such $delta$ we will get a point $x_n$, such that $|x_n - x_0| < delta_n = frac 1n$ but $|f(x_n) - L| geq epsilon_0$. (We don't care about the value of $epsilon_0$ : it is positive, that is what matters).
Question : does $x_n to x_0$? The answer is of course, since we are choosing $x_n$ so that $|x_n - x_0| < frac 1n$, so from the squeeze theorem we get $0 leq |x_n - x_0| < frac 1n$ telling us that $x_n - x_0 to 0$ or that $x_n to x_0$.
Question : does $f(x_n) to L$? The answer is no. The reason is, suppose it were true. Then for all $epsilon > 0$ an $N in mathbb N$ should exist such that whenever $m > N$ we should have $|f(x_m) - L| < epsilon$. But then, from the way we chose $x_m$, we know that $|f(x_m) - L| > epsilon_0$ for all $m$. So for any $epsilon < epsilon_0$, we cannot find any working $delta$, hence convergence is not possible.
This completes the proof.
With this machinery, we may "lift" sequence results to function results.
Proposition : For $c neq 0$ , $lim_{x to x_0} (cf)(x) = c times lim_{x to x_0} f(x)$ if the RHS exists. (For $c = 0$, the LHS exists even if the RHS does not).
Proof : We want to show the LHS exists. Call $lim_{x to x_0}f(x) =L$. From SFR, the LHS is equivalent to : for every sequence $x_n to x_0$, we have to show $(cf)(x_n) to cL$. But then , from SFR on the RHS, certainly $f(x_n) to L$. From what we know about sequences, just multiplying by $c$ gives $(cf)(x_n) to cL$. So the RHS exists and equals $cL$.
Squeeze Theorem : let $g(x) leq f(x) leq h(x)$ be functions, and $x_0,L$ be such that $lim_{x to x_0}g(x) = L$ and $lim_{x to x_0} h(x) = L$. Then, $lim_{x to x_0} f(x) = L$.
Proof : From SFR, we need to show that for any sequence $x_n to x_0$ we have $f(x_n) to L$. But then, $g(x_n) leq f(x_n) leq h(x_n)$ for each $n$, and by SFR on $g$ and $h$, we see that $g(x_n) to L$ and $h(x_n) to L$. Therefore, the squeeze theorem for sequences gives $f(x_n) to L$, as desired.
There are many more properties that one can lift using SFR. Also, SFR applies in more general situations : I just took $mathbb R$ for illustration, but in some more general situations, when we define what $x_n to x_0$ and $lim_{x to x_0} f(x)$ mean, we will be able to provide statements like SFR , which will "lift" sequence properties to function properties nicely. Also, this lifting does not happen in all cases, but that is for another day and date.
answered Feb 11 at 6:57
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.1k33477
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$begingroup$
We go into some generality, when we approach what I would call the link between sequence and function limits.
Sequence limits are given by the $epsilon-N$ definition and function limits by the $epsilon-delta$ definition, if you recall.
Well, first, on the real line, let us prove the following proposition (I am calling this SFR, but you can give it any name you like : it is just nice):
SFR (Sequence-Function Relation) : Let $f : mathbb R to mathbb R$ and $x_0,L in mathbb R$. The following are equivalent :
$(1)displaystylelim_{x to x_0} f(x) = L$.
$(2)$ For every sequence ${x_n} subset mathbb R$ such that $x_n$ converges to $x_0$, we have that the sequence ${f(x_n)}$ converges to $L$.
Notation : $a_n to b$ means "the sequence $a_n$ converges to $b$".
Proof : Suppose $lim_{x to x_0} f(x) = L$. Fix a sequence $y_n to x_0$. We want to show that $f(y_n) to L$. For this , we start with $epsilon > 0$.
For that $epsilon$, we get a $delta > 0$ from $(1)$ such that whenever $|x_0-z| < delta$ we have $|L-f(z)| < epsilon$.
Now, for this $delta > 0$ we get from the fact that $y_n to x_0$, an $N in mathbb N$ such that $m > N implies |y_n -x_0| < delta$.
Therefore :
$$
m > N implies |y_n - x_0| < delta implies |f(y_n) - f(x_0)| < epsilon
$$
and hence $f(y_n) to f(x_0)$.
For the other way, we assume that $(1)$ is not true. So $lim_{x to x_0} f(x) neq L$. We now have to find a sequence $y_n to x_0$ such that $f(y_n) not to L$.
For this, we "negate" the definition of the limit existing : since the limit says that for all $epsilon > 0$ something happens, if the limit does not exist then for some $epsilon_0 > 0$ that something does not happen.
What does not happen? "I can find a $delta > 0$ so that ... happens" does not happen. In other words, you can't find a $delta > 0$ for this $epsilon_0$. Or, every $delta > 0$ fails for this $epsilon_0$.
And what does $delta$ being a failure mean? It means that for some point $x$ such that $|x-x_0| < delta$, we have $|f(x) - L|$ is not less than $epsilon_0$, or in other words $|f(x) - L| geq epsilon_0$.
So, $lim_{x to x_0} f(x) neq L$ means exactly this :
there exists $epsilon_0 > 0$ such that for all $delta > 0$, there exists $x$ such that $|x - x_0| < delta$ but $|f(x) - L| geq epsilon$.
Geometrically, this means : there is an $epsilon_0 >0$ such that you can find points as close to $x_0$ as you want, with the function value at that point at least $epsilon_0$ away from $L$.
With these tools, in the boxed quotes we take $delta = frac 1n$. For each such $delta$ we will get a point $x_n$, such that $|x_n - x_0| < delta_n = frac 1n$ but $|f(x_n) - L| geq epsilon_0$. (We don't care about the value of $epsilon_0$ : it is positive, that is what matters).
Question : does $x_n to x_0$? The answer is of course, since we are choosing $x_n$ so that $|x_n - x_0| < frac 1n$, so from the squeeze theorem we get $0 leq |x_n - x_0| < frac 1n$ telling us that $x_n - x_0 to 0$ or that $x_n to x_0$.
Question : does $f(x_n) to L$? The answer is no. The reason is, suppose it were true. Then for all $epsilon > 0$ an $N in mathbb N$ should exist such that whenever $m > N$ we should have $|f(x_m) - L| < epsilon$. But then, from the way we chose $x_m$, we know that $|f(x_m) - L| > epsilon_0$ for all $m$. So for any $epsilon < epsilon_0$, we cannot find any working $delta$, hence convergence is not possible.
This completes the proof.
With this machinery, we may "lift" sequence results to function results.
Proposition : For $c neq 0$ , $lim_{x to x_0} (cf)(x) = c times lim_{x to x_0} f(x)$ if the RHS exists. (For $c = 0$, the LHS exists even if the RHS does not).
Proof : We want to show the LHS exists. Call $lim_{x to x_0}f(x) =L$. From SFR, the LHS is equivalent to : for every sequence $x_n to x_0$, we have to show $(cf)(x_n) to cL$. But then , from SFR on the RHS, certainly $f(x_n) to L$. From what we know about sequences, just multiplying by $c$ gives $(cf)(x_n) to cL$. So the RHS exists and equals $cL$.
Squeeze Theorem : let $g(x) leq f(x) leq h(x)$ be functions, and $x_0,L$ be such that $lim_{x to x_0}g(x) = L$ and $lim_{x to x_0} h(x) = L$. Then, $lim_{x to x_0} f(x) = L$.
Proof : From SFR, we need to show that for any sequence $x_n to x_0$ we have $f(x_n) to L$. But then, $g(x_n) leq f(x_n) leq h(x_n)$ for each $n$, and by SFR on $g$ and $h$, we see that $g(x_n) to L$ and $h(x_n) to L$. Therefore, the squeeze theorem for sequences gives $f(x_n) to L$, as desired.
There are many more properties that one can lift using SFR. Also, SFR applies in more general situations : I just took $mathbb R$ for illustration, but in some more general situations, when we define what $x_n to x_0$ and $lim_{x to x_0} f(x)$ mean, we will be able to provide statements like SFR , which will "lift" sequence properties to function properties nicely. Also, this lifting does not happen in all cases, but that is for another day and date.
answered Feb 11 at 6:57
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.1k33477
$endgroup$
add a comment |
$begingroup$
We go into some generality, when we approach what I would call the link between sequence and function limits.
Sequence limits are given by the $epsilon-N$ definition and function limits by the $epsilon-delta$ definition, if you recall.
Well, first, on the real line, let us prove the following proposition (I am calling this SFR, but you can give it any name you like : it is just nice):
SFR (Sequence-Function Relation) : Let $f : mathbb R to mathbb R$ and $x_0,L in mathbb R$. The following are equivalent :
$(1)displaystylelim_{x to x_0} f(x) = L$.
$(2)$ For every sequence ${x_n} subset mathbb R$ such that $x_n$ converges to $x_0$, we have that the sequence ${f(x_n)}$ converges to $L$.
Notation : $a_n to b$ means "the sequence $a_n$ converges to $b$".
Proof : Suppose $lim_{x to x_0} f(x) = L$. Fix a sequence $y_n to x_0$. We want to show that $f(y_n) to L$. For this , we start with $epsilon > 0$.
For that $epsilon$, we get a $delta > 0$ from $(1)$ such that whenever $|x_0-z| < delta$ we have $|L-f(z)| < epsilon$.
Now, for this $delta > 0$ we get from the fact that $y_n to x_0$, an $N in mathbb N$ such that $m > N implies |y_n -x_0| < delta$.
Therefore :
$$
m > N implies |y_n - x_0| < delta implies |f(y_n) - f(x_0)| < epsilon
$$
and hence $f(y_n) to f(x_0)$.
For the other way, we assume that $(1)$ is not true. So $lim_{x to x_0} f(x) neq L$. We now have to find a sequence $y_n to x_0$ such that $f(y_n) not to L$.
For this, we "negate" the definition of the limit existing : since the limit says that for all $epsilon > 0$ something happens, if the limit does not exist then for some $epsilon_0 > 0$ that something does not happen.
What does not happen? "I can find a $delta > 0$ so that ... happens" does not happen. In other words, you can't find a $delta > 0$ for this $epsilon_0$. Or, every $delta > 0$ fails for this $epsilon_0$.
And what does $delta$ being a failure mean? It means that for some point $x$ such that $|x-x_0| < delta$, we have $|f(x) - L|$ is not less than $epsilon_0$, or in other words $|f(x) - L| geq epsilon_0$.
So, $lim_{x to x_0} f(x) neq L$ means exactly this :
there exists $epsilon_0 > 0$ such that for all $delta > 0$, there exists $x$ such that $|x - x_0| < delta$ but $|f(x) - L| geq epsilon$.
Geometrically, this means : there is an $epsilon_0 >0$ such that you can find points as close to $x_0$ as you want, with the function value at that point at least $epsilon_0$ away from $L$.
With these tools, in the boxed quotes we take $delta = frac 1n$. For each such $delta$ we will get a point $x_n$, such that $|x_n - x_0| < delta_n = frac 1n$ but $|f(x_n) - L| geq epsilon_0$. (We don't care about the value of $epsilon_0$ : it is positive, that is what matters).
Question : does $x_n to x_0$? The answer is of course, since we are choosing $x_n$ so that $|x_n - x_0| < frac 1n$, so from the squeeze theorem we get $0 leq |x_n - x_0| < frac 1n$ telling us that $x_n - x_0 to 0$ or that $x_n to x_0$.
Question : does $f(x_n) to L$? The answer is no. The reason is, suppose it were true. Then for all $epsilon > 0$ an $N in mathbb N$ should exist such that whenever $m > N$ we should have $|f(x_m) - L| < epsilon$. But then, from the way we chose $x_m$, we know that $|f(x_m) - L| > epsilon_0$ for all $m$. So for any $epsilon < epsilon_0$, we cannot find any working $delta$, hence convergence is not possible.
This completes the proof.
With this machinery, we may "lift" sequence results to function results.
Proposition : For $c neq 0$ , $lim_{x to x_0} (cf)(x) = c times lim_{x to x_0} f(x)$ if the RHS exists. (For $c = 0$, the LHS exists even if the RHS does not).
Proof : We want to show the LHS exists. Call $lim_{x to x_0}f(x) =L$. From SFR, the LHS is equivalent to : for every sequence $x_n to x_0$, we have to show $(cf)(x_n) to cL$. But then , from SFR on the RHS, certainly $f(x_n) to L$. From what we know about sequences, just multiplying by $c$ gives $(cf)(x_n) to cL$. So the RHS exists and equals $cL$.
Squeeze Theorem : let $g(x) leq f(x) leq h(x)$ be functions, and $x_0,L$ be such that $lim_{x to x_0}g(x) = L$ and $lim_{x to x_0} h(x) = L$. Then, $lim_{x to x_0} f(x) = L$.
Proof : From SFR, we need to show that for any sequence $x_n to x_0$ we have $f(x_n) to L$. But then, $g(x_n) leq f(x_n) leq h(x_n)$ for each $n$, and by SFR on $g$ and $h$, we see that $g(x_n) to L$ and $h(x_n) to L$. Therefore, the squeeze theorem for sequences gives $f(x_n) to L$, as desired.
There are many more properties that one can lift using SFR. Also, SFR applies in more general situations : I just took $mathbb R$ for illustration, but in some more general situations, when we define what $x_n to x_0$ and $lim_{x to x_0} f(x)$ mean, we will be able to provide statements like SFR , which will "lift" sequence properties to function properties nicely. Also, this lifting does not happen in all cases, but that is for another day and date.
answered Feb 11 at 6:57
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.1k33477
$endgroup$
add a comment |
$begingroup$
We go into some generality, when we approach what I would call the link between sequence and function limits.
Sequence limits are given by the $epsilon-N$ definition and function limits by the $epsilon-delta$ definition, if you recall.
Well, first, on the real line, let us prove the following proposition (I am calling this SFR, but you can give it any name you like : it is just nice):
SFR (Sequence-Function Relation) : Let $f : mathbb R to mathbb R$ and $x_0,L in mathbb R$. The following are equivalent :
$(1)displaystylelim_{x to x_0} f(x) = L$.
$(2)$ For every sequence ${x_n} subset mathbb R$ such that $x_n$ converges to $x_0$, we have that the sequence ${f(x_n)}$ converges to $L$.
Notation : $a_n to b$ means "the sequence $a_n$ converges to $b$".
Proof : Suppose $lim_{x to x_0} f(x) = L$. Fix a sequence $y_n to x_0$. We want to show that $f(y_n) to L$. For this , we start with $epsilon > 0$.
For that $epsilon$, we get a $delta > 0$ from $(1)$ such that whenever $|x_0-z| < delta$ we have $|L-f(z)| < epsilon$.
Now, for this $delta > 0$ we get from the fact that $y_n to x_0$, an $N in mathbb N$ such that $m > N implies |y_n -x_0| < delta$.
Therefore :
$$
m > N implies |y_n - x_0| < delta implies |f(y_n) - f(x_0)| < epsilon
$$
and hence $f(y_n) to f(x_0)$.
For the other way, we assume that $(1)$ is not true. So $lim_{x to x_0} f(x) neq L$. We now have to find a sequence $y_n to x_0$ such that $f(y_n) not to L$.
For this, we "negate" the definition of the limit existing : since the limit says that for all $epsilon > 0$ something happens, if the limit does not exist then for some $epsilon_0 > 0$ that something does not happen.
What does not happen? "I can find a $delta > 0$ so that ... happens" does not happen. In other words, you can't find a $delta > 0$ for this $epsilon_0$. Or, every $delta > 0$ fails for this $epsilon_0$.
And what does $delta$ being a failure mean? It means that for some point $x$ such that $|x-x_0| < delta$, we have $|f(x) - L|$ is not less than $epsilon_0$, or in other words $|f(x) - L| geq epsilon_0$.
So, $lim_{x to x_0} f(x) neq L$ means exactly this :
there exists $epsilon_0 > 0$ such that for all $delta > 0$, there exists $x$ such that $|x - x_0| < delta$ but $|f(x) - L| geq epsilon$.
Geometrically, this means : there is an $epsilon_0 >0$ such that you can find points as close to $x_0$ as you want, with the function value at that point at least $epsilon_0$ away from $L$.
With these tools, in the boxed quotes we take $delta = frac 1n$. For each such $delta$ we will get a point $x_n$, such that $|x_n - x_0| < delta_n = frac 1n$ but $|f(x_n) - L| geq epsilon_0$. (We don't care about the value of $epsilon_0$ : it is positive, that is what matters).
Question : does $x_n to x_0$? The answer is of course, since we are choosing $x_n$ so that $|x_n - x_0| < frac 1n$, so from the squeeze theorem we get $0 leq |x_n - x_0| < frac 1n$ telling us that $x_n - x_0 to 0$ or that $x_n to x_0$.
Question : does $f(x_n) to L$? The answer is no. The reason is, suppose it were true. Then for all $epsilon > 0$ an $N in mathbb N$ should exist such that whenever $m > N$ we should have $|f(x_m) - L| < epsilon$. But then, from the way we chose $x_m$, we know that $|f(x_m) - L| > epsilon_0$ for all $m$. So for any $epsilon < epsilon_0$, we cannot find any working $delta$, hence convergence is not possible.
This completes the proof.
With this machinery, we may "lift" sequence results to function results.
Proposition : For $c neq 0$ , $lim_{x to x_0} (cf)(x) = c times lim_{x to x_0} f(x)$ if the RHS exists. (For $c = 0$, the LHS exists even if the RHS does not).
Proof : We want to show the LHS exists. Call $lim_{x to x_0}f(x) =L$. From SFR, the LHS is equivalent to : for every sequence $x_n to x_0$, we have to show $(cf)(x_n) to cL$. But then , from SFR on the RHS, certainly $f(x_n) to L$. From what we know about sequences, just multiplying by $c$ gives $(cf)(x_n) to cL$. So the RHS exists and equals $cL$.
Squeeze Theorem : let $g(x) leq f(x) leq h(x)$ be functions, and $x_0,L$ be such that $lim_{x to x_0}g(x) = L$ and $lim_{x to x_0} h(x) = L$. Then, $lim_{x to x_0} f(x) = L$.
Proof : From SFR, we need to show that for any sequence $x_n to x_0$ we have $f(x_n) to L$. But then, $g(x_n) leq f(x_n) leq h(x_n)$ for each $n$, and by SFR on $g$ and $h$, we see that $g(x_n) to L$ and $h(x_n) to L$. Therefore, the squeeze theorem for sequences gives $f(x_n) to L$, as desired.
There are many more properties that one can lift using SFR. Also, SFR applies in more general situations : I just took $mathbb R$ for illustration, but in some more general situations, when we define what $x_n to x_0$ and $lim_{x to x_0} f(x)$ mean, we will be able to provide statements like SFR , which will "lift" sequence properties to function properties nicely. Also, this lifting does not happen in all cases, but that is for another day and date.
answered Feb 11 at 6:57
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.1k33477
$endgroup$
We go into some generality, when we approach what I would call the link between sequence and function limits.
Sequence limits are given by the $epsilon-N$ definition and function limits by the $epsilon-delta$ definition, if you recall.
Well, first, on the real line, let us prove the following proposition (I am calling this SFR, but you can give it any name you like : it is just nice):
SFR (Sequence-Function Relation) : Let $f : mathbb R to mathbb R$ and $x_0,L in mathbb R$. The following are equivalent :
$(1)displaystylelim_{x to x_0} f(x) = L$.
$(2)$ For every sequence ${x_n} subset mathbb R$ such that $x_n$ converges to $x_0$, we have that the sequence ${f(x_n)}$ converges to $L$.
Notation : $a_n to b$ means "the sequence $a_n$ converges to $b$".
Proof : Suppose $lim_{x to x_0} f(x) = L$. Fix a sequence $y_n to x_0$. We want to show that $f(y_n) to L$. For this , we start with $epsilon > 0$.
For that $epsilon$, we get a $delta > 0$ from $(1)$ such that whenever $|x_0-z| < delta$ we have $|L-f(z)| < epsilon$.
Now, for this $delta > 0$ we get from the fact that $y_n to x_0$, an $N in mathbb N$ such that $m > N implies |y_n -x_0| < delta$.
Therefore :
$$
m > N implies |y_n - x_0| < delta implies |f(y_n) - f(x_0)| < epsilon
$$
and hence $f(y_n) to f(x_0)$.
For the other way, we assume that $(1)$ is not true. So $lim_{x to x_0} f(x) neq L$. We now have to find a sequence $y_n to x_0$ such that $f(y_n) not to L$.
For this, we "negate" the definition of the limit existing : since the limit says that for all $epsilon > 0$ something happens, if the limit does not exist then for some $epsilon_0 > 0$ that something does not happen.
What does not happen? "I can find a $delta > 0$ so that ... happens" does not happen. In other words, you can't find a $delta > 0$ for this $epsilon_0$. Or, every $delta > 0$ fails for this $epsilon_0$.
And what does $delta$ being a failure mean? It means that for some point $x$ such that $|x-x_0| < delta$, we have $|f(x) - L|$ is not less than $epsilon_0$, or in other words $|f(x) - L| geq epsilon_0$.
So, $lim_{x to x_0} f(x) neq L$ means exactly this :
there exists $epsilon_0 > 0$ such that for all $delta > 0$, there exists $x$ such that $|x - x_0| < delta$ but $|f(x) - L| geq epsilon$.
Geometrically, this means : there is an $epsilon_0 >0$ such that you can find points as close to $x_0$ as you want, with the function value at that point at least $epsilon_0$ away from $L$.
With these tools, in the boxed quotes we take $delta = frac 1n$. For each such $delta$ we will get a point $x_n$, such that $|x_n - x_0| < delta_n = frac 1n$ but $|f(x_n) - L| geq epsilon_0$. (We don't care about the value of $epsilon_0$ : it is positive, that is what matters).
Question : does $x_n to x_0$? The answer is of course, since we are choosing $x_n$ so that $|x_n - x_0| < frac 1n$, so from the squeeze theorem we get $0 leq |x_n - x_0| < frac 1n$ telling us that $x_n - x_0 to 0$ or that $x_n to x_0$.
Question : does $f(x_n) to L$? The answer is no. The reason is, suppose it were true. Then for all $epsilon > 0$ an $N in mathbb N$ should exist such that whenever $m > N$ we should have $|f(x_m) - L| < epsilon$. But then, from the way we chose $x_m$, we know that $|f(x_m) - L| > epsilon_0$ for all $m$. So for any $epsilon < epsilon_0$, we cannot find any working $delta$, hence convergence is not possible.
This completes the proof.
With this machinery, we may "lift" sequence results to function results.
Proposition : For $c neq 0$ , $lim_{x to x_0} (cf)(x) = c times lim_{x to x_0} f(x)$ if the RHS exists. (For $c = 0$, the LHS exists even if the RHS does not).
Proof : We want to show the LHS exists. Call $lim_{x to x_0}f(x) =L$. From SFR, the LHS is equivalent to : for every sequence $x_n to x_0$, we have to show $(cf)(x_n) to cL$. But then , from SFR on the RHS, certainly $f(x_n) to L$. From what we know about sequences, just multiplying by $c$ gives $(cf)(x_n) to cL$. So the RHS exists and equals $cL$.
Squeeze Theorem : let $g(x) leq f(x) leq h(x)$ be functions, and $x_0,L$ be such that $lim_{x to x_0}g(x) = L$ and $lim_{x to x_0} h(x) = L$. Then, $lim_{x to x_0} f(x) = L$.
Proof : From SFR, we need to show that for any sequence $x_n to x_0$ we have $f(x_n) to L$. But then, $g(x_n) leq f(x_n) leq h(x_n)$ for each $n$, and by SFR on $g$ and $h$, we see that $g(x_n) to L$ and $h(x_n) to L$. Therefore, the squeeze theorem for sequences gives $f(x_n) to L$, as desired.
There are many more properties that one can lift using SFR. Also, SFR applies in more general situations : I just took $mathbb R$ for illustration, but in some more general situations, when we define what $x_n to x_0$ and $lim_{x to x_0} f(x)$ mean, we will be able to provide statements like SFR , which will "lift" sequence properties to function properties nicely. Also, this lifting does not happen in all cases, but that is for another day and date.
answered Feb 11 at 6:57
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.1k33477
answered Feb 11 at 6:57
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.1k33477
answered Feb 11 at 6:57
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.1k33477
answered Feb 11 at 6:57
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.1k33477
39.1k33477
add a comment |
add a comment |
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$begingroup$
This is intuitively clear if you think about limits in terms of arbitrarily close approximations. If you look at the parallel proofs using the definitions of each kind of limit you will see the same essential arguments.
$endgroup$
– Ethan Bolker
Jan 23 at 12:53
$begingroup$
If you are expecting a proof then, have a look at this: en.wikipedia.org/wiki/Squeeze_theorem. For the second proof just use the $epsilon-N$ definition of limits.
$endgroup$
– Sujit Bhattacharyya
Jan 23 at 12:58
$begingroup$
The definition of limit of sequences and limit of functions are very similar and hence one expects similar results. To be more formal, if you understand the proof for these results on sequences you should not have any trouble writing a proof for corresponding result on functions.
$endgroup$
– Paramanand Singh
Jan 24 at 3:28