Given $lim_{ntoinfty}x_n = x ne 0$ and $lim_{ntoinfty}y_n = pminfty$ show that $lim x_ny_n = mpinfty$ when...
$begingroup$
Given
$$
begin{cases}
lim_{ntoinfty}x_n = x ne 0 \
lim_{ntoinfty}y_n = pminfty
end{cases}
$$
Show that for $x < 0$:
$$
lim x_ny_n = mpinfty
$$
I've started with the case when $lim y_n = +infty$:
$$
forall varepsilon >0 exists N_1 in Bbb N:forall n>N_1 implies y_n > varepsilon
$$
On the other hand the definition for $lim x_n$ is going to be:
$$
forall varepsilon > 0 exists N_2 in Bbb N: forall n > N_2 implies |x_n - x| < varepsilon
$$
That means we may choose some $N$ starting from which the following is true:
$$
forall varepsilon > 0 exists N = max{N_1, N_2}: forall n> N implies y_n > varepsilon text{and} |x_n - x| < varepsilon
$$
So having that in mind we may consider the following system:
$$
begin{cases}
|x_n - x| < varepsilon \
y_n > varepsilon
end{cases} iff
begin{cases}
begin{align}
-varepsilon < &x_n -x < varepsilon \
-&y_n < -varepsilon
end{align}
end{cases}
$$
So now if we multiply the inequalities one may obtain:
$$
-y_n(x_n - x) < -varepsilon^2 iff -x_ny_n + y_nx < varepsilon^2 iff x_ny_n>-varepsilon^2 + y_nx
$$
At this point I got stuck, basically my idea was to utilize the definition of limits and then combine the two cases to arrive at a definition of a limit but for the sequence $x_ny_n$, but not sure how to proceed.
What steps should I take to prove what's in the problem section?
calculus sequences-and-series limits
asked Dec 9 '18 at 14:02
romanroman
2,11521222
$endgroup$
add a comment |
$begingroup$
Given
$$
begin{cases}
lim_{ntoinfty}x_n = x ne 0 \
lim_{ntoinfty}y_n = pminfty
end{cases}
$$
Show that for $x < 0$:
$$
lim x_ny_n = mpinfty
$$
I've started with the case when $lim y_n = +infty$:
$$
forall varepsilon >0 exists N_1 in Bbb N:forall n>N_1 implies y_n > varepsilon
$$
On the other hand the definition for $lim x_n$ is going to be:
$$
forall varepsilon > 0 exists N_2 in Bbb N: forall n > N_2 implies |x_n - x| < varepsilon
$$
That means we may choose some $N$ starting from which the following is true:
$$
forall varepsilon > 0 exists N = max{N_1, N_2}: forall n> N implies y_n > varepsilon text{and} |x_n - x| < varepsilon
$$
So having that in mind we may consider the following system:
$$
begin{cases}
|x_n - x| < varepsilon \
y_n > varepsilon
end{cases} iff
begin{cases}
begin{align}
-varepsilon < &x_n -x < varepsilon \
-&y_n < -varepsilon
end{align}
end{cases}
$$
So now if we multiply the inequalities one may obtain:
$$
-y_n(x_n - x) < -varepsilon^2 iff -x_ny_n + y_nx < varepsilon^2 iff x_ny_n>-varepsilon^2 + y_nx
$$
At this point I got stuck, basically my idea was to utilize the definition of limits and then combine the two cases to arrive at a definition of a limit but for the sequence $x_ny_n$, but not sure how to proceed.
What steps should I take to prove what's in the problem section?
calculus sequences-and-series limits
asked Dec 9 '18 at 14:02
romanroman
2,11521222
$endgroup$
$begingroup$
Is the $varepsilon-delta$ way obligatory?
$endgroup$
– Rebellos
Dec 9 '18 at 14:09
$begingroup$
@Rebellos not necessarily, that was just my try
$endgroup$
– roman
Dec 9 '18 at 14:10
add a comment |
$begingroup$
Given
$$
begin{cases}
lim_{ntoinfty}x_n = x ne 0 \
lim_{ntoinfty}y_n = pminfty
end{cases}
$$
Show that for $x < 0$:
$$
lim x_ny_n = mpinfty
$$
I've started with the case when $lim y_n = +infty$:
$$
forall varepsilon >0 exists N_1 in Bbb N:forall n>N_1 implies y_n > varepsilon
$$
On the other hand the definition for $lim x_n$ is going to be:
$$
forall varepsilon > 0 exists N_2 in Bbb N: forall n > N_2 implies |x_n - x| < varepsilon
$$
That means we may choose some $N$ starting from which the following is true:
$$
forall varepsilon > 0 exists N = max{N_1, N_2}: forall n> N implies y_n > varepsilon text{and} |x_n - x| < varepsilon
$$
So having that in mind we may consider the following system:
$$
begin{cases}
|x_n - x| < varepsilon \
y_n > varepsilon
end{cases} iff
begin{cases}
begin{align}
-varepsilon < &x_n -x < varepsilon \
-&y_n < -varepsilon
end{align}
end{cases}
$$
So now if we multiply the inequalities one may obtain:
$$
-y_n(x_n - x) < -varepsilon^2 iff -x_ny_n + y_nx < varepsilon^2 iff x_ny_n>-varepsilon^2 + y_nx
$$
At this point I got stuck, basically my idea was to utilize the definition of limits and then combine the two cases to arrive at a definition of a limit but for the sequence $x_ny_n$, but not sure how to proceed.
What steps should I take to prove what's in the problem section?
calculus sequences-and-series limits
asked Dec 9 '18 at 14:02
romanroman
2,11521222
$endgroup$
Given
$$
begin{cases}
lim_{ntoinfty}x_n = x ne 0 \
lim_{ntoinfty}y_n = pminfty
end{cases}
$$
Show that for $x < 0$:
$$
lim x_ny_n = mpinfty
$$
I've started with the case when $lim y_n = +infty$:
$$
forall varepsilon >0 exists N_1 in Bbb N:forall n>N_1 implies y_n > varepsilon
$$
On the other hand the definition for $lim x_n$ is going to be:
$$
forall varepsilon > 0 exists N_2 in Bbb N: forall n > N_2 implies |x_n - x| < varepsilon
$$
That means we may choose some $N$ starting from which the following is true:
$$
forall varepsilon > 0 exists N = max{N_1, N_2}: forall n> N implies y_n > varepsilon text{and} |x_n - x| < varepsilon
$$
So having that in mind we may consider the following system:
$$
begin{cases}
|x_n - x| < varepsilon \
y_n > varepsilon
end{cases} iff
begin{cases}
begin{align}
-varepsilon < &x_n -x < varepsilon \
-&y_n < -varepsilon
end{align}
end{cases}
$$
So now if we multiply the inequalities one may obtain:
$$
-y_n(x_n - x) < -varepsilon^2 iff -x_ny_n + y_nx < varepsilon^2 iff x_ny_n>-varepsilon^2 + y_nx
$$
At this point I got stuck, basically my idea was to utilize the definition of limits and then combine the two cases to arrive at a definition of a limit but for the sequence $x_ny_n$, but not sure how to proceed.
What steps should I take to prove what's in the problem section?
calculus sequences-and-series limits
calculus sequences-and-series limits
asked Dec 9 '18 at 14:02
romanroman
2,11521222
asked Dec 9 '18 at 14:02
romanroman
2,11521222
edited Dec 9 '18 at 14:10
roman
asked Dec 9 '18 at 14:02
romanroman
2,11521222
asked Dec 9 '18 at 14:02
romanroman
2,11521222
asked Dec 9 '18 at 14:02
romanroman
2,11521222
2,11521222
$begingroup$
Is the $varepsilon-delta$ way obligatory?
$endgroup$
– Rebellos
Dec 9 '18 at 14:09
$begingroup$
@Rebellos not necessarily, that was just my try
$endgroup$
– roman
Dec 9 '18 at 14:10
add a comment |
$begingroup$
Is the $varepsilon-delta$ way obligatory?
$endgroup$
– Rebellos
Dec 9 '18 at 14:09
$begingroup$
@Rebellos not necessarily, that was just my try
$endgroup$
– roman
Dec 9 '18 at 14:10
$begingroup$
Is the $varepsilon-delta$ way obligatory?
$endgroup$
– Rebellos
Dec 9 '18 at 14:09
$begingroup$
Is the $varepsilon-delta$ way obligatory?
$endgroup$
– Rebellos
Dec 9 '18 at 14:09
$begingroup$
@Rebellos not necessarily, that was just my try
$endgroup$
– roman
Dec 9 '18 at 14:10
$begingroup$
@Rebellos not necessarily, that was just my try
$endgroup$
– roman
Dec 9 '18 at 14:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose $lim y_n = +infty$:
We want to show that for any $M>0$, we can find $N>0$, such that if $n>N$, then $x_ny_n < -M$.
We know that
$$
forall M >0 exists N_{1,M} in Bbb N:forall n>N_{1,M} implies y_n > M
$$
In paticular,
$$exists N_{1,frac{2M}{|x|}} in Bbb N:forall n>N_{1,frac{2M}{|x|}} implies y_n > frac{2M}{|x|} iff (-x)y_n > 2M iff xy_n < -2M
$$
On the other hand the definition for $lim x_n$ is going to be:
$$
forall varepsilon > 0 exists N_{2,varepsilon} in Bbb N: forall n > N_{2,varepsilon} implies |x_n - x| < varepsilon
$$
In particular, $$ exists N_{2,-frac{x}2} in Bbb N: forall n > N_{2,-frac{x}2} implies |x_n - x| < -frac{x}2 implies x_n < frac{x}2.
$$
Hence for any $n> maxleft(N_{1,frac{2M}{|x|}} , N_{2,-frac{x}2}right), x_ny_n < frac{x}{2}y_n < frac{(-2M)}{2}=-M$
answered Dec 9 '18 at 14:19
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
$begingroup$
Could you please elaborate on the parts starting with "In particular, "?
$endgroup$
– roman
Dec 9 '18 at 14:31
$begingroup$
I choose my particular $M$ to be $frac{2M}{|x|}$. and for the seocnd part, I choose my particular $varepsilon$ to be $-frac{x}2$
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 14:33
$begingroup$
Oh, i see now, thanks.
$endgroup$
– roman
Dec 9 '18 at 14:35
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3032404%2fgiven-lim-n-to-inftyx-n-x-ne-0-and-lim-n-to-inftyy-n-pm-infty-sh%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose $lim y_n = +infty$:
We want to show that for any $M>0$, we can find $N>0$, such that if $n>N$, then $x_ny_n < -M$.
We know that
$$
forall M >0 exists N_{1,M} in Bbb N:forall n>N_{1,M} implies y_n > M
$$
In paticular,
$$exists N_{1,frac{2M}{|x|}} in Bbb N:forall n>N_{1,frac{2M}{|x|}} implies y_n > frac{2M}{|x|} iff (-x)y_n > 2M iff xy_n < -2M
$$
On the other hand the definition for $lim x_n$ is going to be:
$$
forall varepsilon > 0 exists N_{2,varepsilon} in Bbb N: forall n > N_{2,varepsilon} implies |x_n - x| < varepsilon
$$
In particular, $$ exists N_{2,-frac{x}2} in Bbb N: forall n > N_{2,-frac{x}2} implies |x_n - x| < -frac{x}2 implies x_n < frac{x}2.
$$
Hence for any $n> maxleft(N_{1,frac{2M}{|x|}} , N_{2,-frac{x}2}right), x_ny_n < frac{x}{2}y_n < frac{(-2M)}{2}=-M$
answered Dec 9 '18 at 14:19
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
$begingroup$
Could you please elaborate on the parts starting with "In particular, "?
$endgroup$
– roman
Dec 9 '18 at 14:31
$begingroup$
I choose my particular $M$ to be $frac{2M}{|x|}$. and for the seocnd part, I choose my particular $varepsilon$ to be $-frac{x}2$
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 14:33
$begingroup$
Oh, i see now, thanks.
$endgroup$
– roman
Dec 9 '18 at 14:35
add a comment |
$begingroup$
Suppose $lim y_n = +infty$:
We want to show that for any $M>0$, we can find $N>0$, such that if $n>N$, then $x_ny_n < -M$.
We know that
$$
forall M >0 exists N_{1,M} in Bbb N:forall n>N_{1,M} implies y_n > M
$$
In paticular,
$$exists N_{1,frac{2M}{|x|}} in Bbb N:forall n>N_{1,frac{2M}{|x|}} implies y_n > frac{2M}{|x|} iff (-x)y_n > 2M iff xy_n < -2M
$$
On the other hand the definition for $lim x_n$ is going to be:
$$
forall varepsilon > 0 exists N_{2,varepsilon} in Bbb N: forall n > N_{2,varepsilon} implies |x_n - x| < varepsilon
$$
In particular, $$ exists N_{2,-frac{x}2} in Bbb N: forall n > N_{2,-frac{x}2} implies |x_n - x| < -frac{x}2 implies x_n < frac{x}2.
$$
Hence for any $n> maxleft(N_{1,frac{2M}{|x|}} , N_{2,-frac{x}2}right), x_ny_n < frac{x}{2}y_n < frac{(-2M)}{2}=-M$
answered Dec 9 '18 at 14:19
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
$begingroup$
Could you please elaborate on the parts starting with "In particular, "?
$endgroup$
– roman
Dec 9 '18 at 14:31
$begingroup$
I choose my particular $M$ to be $frac{2M}{|x|}$. and for the seocnd part, I choose my particular $varepsilon$ to be $-frac{x}2$
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 14:33
$begingroup$
Oh, i see now, thanks.
$endgroup$
– roman
Dec 9 '18 at 14:35
add a comment |
$begingroup$
Suppose $lim y_n = +infty$:
We want to show that for any $M>0$, we can find $N>0$, such that if $n>N$, then $x_ny_n < -M$.
We know that
$$
forall M >0 exists N_{1,M} in Bbb N:forall n>N_{1,M} implies y_n > M
$$
In paticular,
$$exists N_{1,frac{2M}{|x|}} in Bbb N:forall n>N_{1,frac{2M}{|x|}} implies y_n > frac{2M}{|x|} iff (-x)y_n > 2M iff xy_n < -2M
$$
On the other hand the definition for $lim x_n$ is going to be:
$$
forall varepsilon > 0 exists N_{2,varepsilon} in Bbb N: forall n > N_{2,varepsilon} implies |x_n - x| < varepsilon
$$
In particular, $$ exists N_{2,-frac{x}2} in Bbb N: forall n > N_{2,-frac{x}2} implies |x_n - x| < -frac{x}2 implies x_n < frac{x}2.
$$
Hence for any $n> maxleft(N_{1,frac{2M}{|x|}} , N_{2,-frac{x}2}right), x_ny_n < frac{x}{2}y_n < frac{(-2M)}{2}=-M$
answered Dec 9 '18 at 14:19
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
Suppose $lim y_n = +infty$:
We want to show that for any $M>0$, we can find $N>0$, such that if $n>N$, then $x_ny_n < -M$.
We know that
$$
forall M >0 exists N_{1,M} in Bbb N:forall n>N_{1,M} implies y_n > M
$$
In paticular,
$$exists N_{1,frac{2M}{|x|}} in Bbb N:forall n>N_{1,frac{2M}{|x|}} implies y_n > frac{2M}{|x|} iff (-x)y_n > 2M iff xy_n < -2M
$$
On the other hand the definition for $lim x_n$ is going to be:
$$
forall varepsilon > 0 exists N_{2,varepsilon} in Bbb N: forall n > N_{2,varepsilon} implies |x_n - x| < varepsilon
$$
In particular, $$ exists N_{2,-frac{x}2} in Bbb N: forall n > N_{2,-frac{x}2} implies |x_n - x| < -frac{x}2 implies x_n < frac{x}2.
$$
Hence for any $n> maxleft(N_{1,frac{2M}{|x|}} , N_{2,-frac{x}2}right), x_ny_n < frac{x}{2}y_n < frac{(-2M)}{2}=-M$
answered Dec 9 '18 at 14:19
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 9 '18 at 14:19
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 9 '18 at 14:19
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 9 '18 at 14:19
Siong Thye GohSiong Thye Goh
100k1466117
100k1466117
$begingroup$
Could you please elaborate on the parts starting with "In particular, "?
$endgroup$
– roman
Dec 9 '18 at 14:31
$begingroup$
I choose my particular $M$ to be $frac{2M}{|x|}$. and for the seocnd part, I choose my particular $varepsilon$ to be $-frac{x}2$
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 14:33
$begingroup$
Oh, i see now, thanks.
$endgroup$
– roman
Dec 9 '18 at 14:35
add a comment |
$begingroup$
Could you please elaborate on the parts starting with "In particular, "?
$endgroup$
– roman
Dec 9 '18 at 14:31
$begingroup$
I choose my particular $M$ to be $frac{2M}{|x|}$. and for the seocnd part, I choose my particular $varepsilon$ to be $-frac{x}2$
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 14:33
$begingroup$
Oh, i see now, thanks.
$endgroup$
– roman
Dec 9 '18 at 14:35
$begingroup$
Could you please elaborate on the parts starting with "In particular, "?
$endgroup$
– roman
Dec 9 '18 at 14:31
$begingroup$
Could you please elaborate on the parts starting with "In particular, "?
$endgroup$
– roman
Dec 9 '18 at 14:31
$begingroup$
I choose my particular $M$ to be $frac{2M}{|x|}$. and for the seocnd part, I choose my particular $varepsilon$ to be $-frac{x}2$
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 14:33
$begingroup$
I choose my particular $M$ to be $frac{2M}{|x|}$. and for the seocnd part, I choose my particular $varepsilon$ to be $-frac{x}2$
$endgroup$
– Siong Thye Goh
Dec 9 '18 at 14:33
$begingroup$
Oh, i see now, thanks.
$endgroup$
– roman
Dec 9 '18 at 14:35
$begingroup$
Oh, i see now, thanks.
$endgroup$
– roman
Dec 9 '18 at 14:35
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3032404%2fgiven-lim-n-to-inftyx-n-x-ne-0-and-lim-n-to-inftyy-n-pm-infty-sh%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Is the $varepsilon-delta$ way obligatory?
$endgroup$
– Rebellos
Dec 9 '18 at 14:09
$begingroup$
@Rebellos not necessarily, that was just my try
$endgroup$
– roman
Dec 9 '18 at 14:10