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
2,17321224
$endgroup$
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
2,17321224
$endgroup$
$begingroup$
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
2,17321224
$endgroup$
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
2,17321224
asked Dec 9 '18 at 14:02
romanroman
2,17321224
edited Dec 9 '18 at 14:10
roman
asked Dec 9 '18 at 14:02
romanroman
2,17321224
asked Dec 9 '18 at 14:02
romanroman
2,17321224
asked Dec 9 '18 at 14:02
romanroman
2,17321224
2,17321224
$begingroup$
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
101k1466117
$endgroup$
$begingroup$
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 |
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1 Answer
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$begingroup$
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
101k1466117
$endgroup$
$begingroup$
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
101k1466117
$endgroup$
$begingroup$
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
101k1466117
$endgroup$
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
101k1466117
answered Dec 9 '18 at 14:19
Siong Thye GohSiong Thye Goh
101k1466117
answered Dec 9 '18 at 14:19
Siong Thye GohSiong Thye Goh
101k1466117
answered Dec 9 '18 at 14:19
Siong Thye GohSiong Thye Goh
101k1466117
101k1466117
$begingroup$
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 |
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$begingroup$
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