Proof verification for $lim_{ntoinfty}(sqrt{n^2-1} - sqrt n) = +infty$
$begingroup$
Show that:
$$
lim_{ntoinfty}left(sqrt{n^2-1} - sqrt nright) = +infty
$$
I've started it this way.
Lemma:
Let $x_n$ and $y_n$ be two sequences. Claim:
If:
$$
begin{cases}
&lim_{ntoinfty} x_n =+infty \
&exists Nin Bbb N, forall n >N:y_nge c > 0
end{cases}
$$
Then:
$$
lim_{ntoinfty}(x_ny_n) = +infty
$$
Proof:
$Box$ Start with definition of limit for this case:
$$
forallvarepsilon>0, exists N_1inBbb N: forall n > N_1 implies x_n >varepsilon
$$
Also:
$$
exists N_2inBbb N:forall n>N_2 implies y_n ge c > 0
$$
Let:
$$
N = max{N_1, N_2}
$$
Then starting from this $N$ we obtain:
$$
x_ncdot y_n > ccdot varepsilon
$$
And we have that:
$$
forallvarepsilon>0, exists N =max{N_1, N_2}inBbb N: forall n > N implies x_n y_n > cvarepsilon
$$
Thus:
$$
lim_{ntoinfty}(x_ny_n) = +infty Box
$$
Now back to the initial problem. Let:
$$
z_n = sqrt{n^2-1} - sqrt n = frac{n^2 - n - 1}{sqrt{n^2 - 1} + sqrt{n}}
$$
Define:
$$
x_n = n - 1 - {1over n} \
y_n = frac{n}{sqrt{n^2 - 1} + sqrt{n}}
$$
Obviously $y_n ge c > 0$ for some $N$ and $n>N$. Also $x_n to +infty$, then by lemma:
$$
lim_{ntoinfty}z_n = lim_{ntoinfty}{x_ny_n} = +infty
$$
I know this is a bit overkill, but i wanted to use that exact lemma for the proof. Apart from that, is it valid?
BTW here is a visualization for $x_n, y_n$
Update
Since it is not clear where the lemma comes from here is the problem from the problem book right before the limit.
Let:
$$
lim_{ntoinfty}x_n = a , text{where} a = +infty text{or} a = -infty
$$
Prove that if for all $n$ starting from some $N$ $y_n ge c > 0$ then
$$
lim_{ntoinfty}x_ny_n = a
$$
And if for all $n$ starting from some $N$ $y_n le c < 0$ then
$$
lim_{ntoinfty}x_ny_n = -a
$$
No other constraints are given.
calculus limits proof-verification epsilon-delta
asked Dec 6 '18 at 18:31
romanroman
2,17321224
$endgroup$
|
show 2 more comments
$begingroup$
Show that:
$$
lim_{ntoinfty}left(sqrt{n^2-1} - sqrt nright) = +infty
$$
I've started it this way.
Lemma:
Let $x_n$ and $y_n$ be two sequences. Claim:
If:
$$
begin{cases}
&lim_{ntoinfty} x_n =+infty \
&exists Nin Bbb N, forall n >N:y_nge c > 0
end{cases}
$$
Then:
$$
lim_{ntoinfty}(x_ny_n) = +infty
$$
Proof:
$Box$ Start with definition of limit for this case:
$$
forallvarepsilon>0, exists N_1inBbb N: forall n > N_1 implies x_n >varepsilon
$$
Also:
$$
exists N_2inBbb N:forall n>N_2 implies y_n ge c > 0
$$
Let:
$$
N = max{N_1, N_2}
$$
Then starting from this $N$ we obtain:
$$
x_ncdot y_n > ccdot varepsilon
$$
And we have that:
$$
forallvarepsilon>0, exists N =max{N_1, N_2}inBbb N: forall n > N implies x_n y_n > cvarepsilon
$$
Thus:
$$
lim_{ntoinfty}(x_ny_n) = +infty Box
$$
Now back to the initial problem. Let:
$$
z_n = sqrt{n^2-1} - sqrt n = frac{n^2 - n - 1}{sqrt{n^2 - 1} + sqrt{n}}
$$
Define:
$$
x_n = n - 1 - {1over n} \
y_n = frac{n}{sqrt{n^2 - 1} + sqrt{n}}
$$
Obviously $y_n ge c > 0$ for some $N$ and $n>N$. Also $x_n to +infty$, then by lemma:
$$
lim_{ntoinfty}z_n = lim_{ntoinfty}{x_ny_n} = +infty
$$
I know this is a bit overkill, but i wanted to use that exact lemma for the proof. Apart from that, is it valid?
BTW here is a visualization for $x_n, y_n$
Update
Since it is not clear where the lemma comes from here is the problem from the problem book right before the limit.
Let:
$$
lim_{ntoinfty}x_n = a , text{where} a = +infty text{or} a = -infty
$$
Prove that if for all $n$ starting from some $N$ $y_n ge c > 0$ then
$$
lim_{ntoinfty}x_ny_n = a
$$
And if for all $n$ starting from some $N$ $y_n le c < 0$ then
$$
lim_{ntoinfty}x_ny_n = -a
$$
No other constraints are given.
calculus limits proof-verification epsilon-delta
asked Dec 6 '18 at 18:31
romanroman
2,17321224
$endgroup$
$begingroup$
What's $c$ in your lemma? It feels like it's just randomly introduced?
$endgroup$
– Jam
Dec 6 '18 at 19:30
$begingroup$
@Jam, the exercise to prove this lemma comes right before the exercise on the limit (which in in the question section). No constraints for $c$ are given except for the fact that from some $N$ $y_n ge c > 0$. This lemma is actually one of the several of the same kind in the exercise before limit.
$endgroup$
– roman
Dec 6 '18 at 19:32
$begingroup$
I may be mistaken but I think you need more constraints on $c$. It seems like there's nothing stopping us from having $c=1/varepsilon>0$, in which case $x_ny_n>cvarepsilon=1$ doesn't tell us much.
$endgroup$
– Jam
Dec 6 '18 at 19:39
$begingroup$
In other words, you've shown that $x$ is unbounded above and $y$ is bounded from below but how do we know that all $y$ have the same lower bound? And how do we know that the lower bound of $y$ doesn't get small very quickly (counteracting the size of $varepsilon$)?
$endgroup$
– Jam
Dec 6 '18 at 19:42
$begingroup$
@Jam, I have updated the question and added more context of where that lemma came from
$endgroup$
– roman
Dec 6 '18 at 22:10
|
show 2 more comments
$begingroup$
Show that:
$$
lim_{ntoinfty}left(sqrt{n^2-1} - sqrt nright) = +infty
$$
I've started it this way.
Lemma:
Let $x_n$ and $y_n$ be two sequences. Claim:
If:
$$
begin{cases}
&lim_{ntoinfty} x_n =+infty \
&exists Nin Bbb N, forall n >N:y_nge c > 0
end{cases}
$$
Then:
$$
lim_{ntoinfty}(x_ny_n) = +infty
$$
Proof:
$Box$ Start with definition of limit for this case:
$$
forallvarepsilon>0, exists N_1inBbb N: forall n > N_1 implies x_n >varepsilon
$$
Also:
$$
exists N_2inBbb N:forall n>N_2 implies y_n ge c > 0
$$
Let:
$$
N = max{N_1, N_2}
$$
Then starting from this $N$ we obtain:
$$
x_ncdot y_n > ccdot varepsilon
$$
And we have that:
$$
forallvarepsilon>0, exists N =max{N_1, N_2}inBbb N: forall n > N implies x_n y_n > cvarepsilon
$$
Thus:
$$
lim_{ntoinfty}(x_ny_n) = +infty Box
$$
Now back to the initial problem. Let:
$$
z_n = sqrt{n^2-1} - sqrt n = frac{n^2 - n - 1}{sqrt{n^2 - 1} + sqrt{n}}
$$
Define:
$$
x_n = n - 1 - {1over n} \
y_n = frac{n}{sqrt{n^2 - 1} + sqrt{n}}
$$
Obviously $y_n ge c > 0$ for some $N$ and $n>N$. Also $x_n to +infty$, then by lemma:
$$
lim_{ntoinfty}z_n = lim_{ntoinfty}{x_ny_n} = +infty
$$
I know this is a bit overkill, but i wanted to use that exact lemma for the proof. Apart from that, is it valid?
BTW here is a visualization for $x_n, y_n$
Update
Since it is not clear where the lemma comes from here is the problem from the problem book right before the limit.
Let:
$$
lim_{ntoinfty}x_n = a , text{where} a = +infty text{or} a = -infty
$$
Prove that if for all $n$ starting from some $N$ $y_n ge c > 0$ then
$$
lim_{ntoinfty}x_ny_n = a
$$
And if for all $n$ starting from some $N$ $y_n le c < 0$ then
$$
lim_{ntoinfty}x_ny_n = -a
$$
No other constraints are given.
calculus limits proof-verification epsilon-delta
asked Dec 6 '18 at 18:31
romanroman
2,17321224
$endgroup$
Show that:
$$
lim_{ntoinfty}left(sqrt{n^2-1} - sqrt nright) = +infty
$$
I've started it this way.
Lemma:
Let $x_n$ and $y_n$ be two sequences. Claim:
If:
$$
begin{cases}
&lim_{ntoinfty} x_n =+infty \
&exists Nin Bbb N, forall n >N:y_nge c > 0
end{cases}
$$
Then:
$$
lim_{ntoinfty}(x_ny_n) = +infty
$$
Proof:
$Box$ Start with definition of limit for this case:
$$
forallvarepsilon>0, exists N_1inBbb N: forall n > N_1 implies x_n >varepsilon
$$
Also:
$$
exists N_2inBbb N:forall n>N_2 implies y_n ge c > 0
$$
Let:
$$
N = max{N_1, N_2}
$$
Then starting from this $N$ we obtain:
$$
x_ncdot y_n > ccdot varepsilon
$$
And we have that:
$$
forallvarepsilon>0, exists N =max{N_1, N_2}inBbb N: forall n > N implies x_n y_n > cvarepsilon
$$
Thus:
$$
lim_{ntoinfty}(x_ny_n) = +infty Box
$$
Now back to the initial problem. Let:
$$
z_n = sqrt{n^2-1} - sqrt n = frac{n^2 - n - 1}{sqrt{n^2 - 1} + sqrt{n}}
$$
Define:
$$
x_n = n - 1 - {1over n} \
y_n = frac{n}{sqrt{n^2 - 1} + sqrt{n}}
$$
Obviously $y_n ge c > 0$ for some $N$ and $n>N$. Also $x_n to +infty$, then by lemma:
$$
lim_{ntoinfty}z_n = lim_{ntoinfty}{x_ny_n} = +infty
$$
I know this is a bit overkill, but i wanted to use that exact lemma for the proof. Apart from that, is it valid?
BTW here is a visualization for $x_n, y_n$
Update
Since it is not clear where the lemma comes from here is the problem from the problem book right before the limit.
Let:
$$
lim_{ntoinfty}x_n = a , text{where} a = +infty text{or} a = -infty
$$
Prove that if for all $n$ starting from some $N$ $y_n ge c > 0$ then
$$
lim_{ntoinfty}x_ny_n = a
$$
And if for all $n$ starting from some $N$ $y_n le c < 0$ then
$$
lim_{ntoinfty}x_ny_n = -a
$$
No other constraints are given.
calculus limits proof-verification epsilon-delta
calculus limits proof-verification epsilon-delta
asked Dec 6 '18 at 18:31
romanroman
2,17321224
asked Dec 6 '18 at 18:31
romanroman
2,17321224
edited Dec 6 '18 at 22:25
roman
asked Dec 6 '18 at 18:31
romanroman
2,17321224
asked Dec 6 '18 at 18:31
romanroman
2,17321224
asked Dec 6 '18 at 18:31
romanroman
2,17321224
2,17321224
$begingroup$
What's $c$ in your lemma? It feels like it's just randomly introduced?
$endgroup$
– Jam
Dec 6 '18 at 19:30
$begingroup$
@Jam, the exercise to prove this lemma comes right before the exercise on the limit (which in in the question section). No constraints for $c$ are given except for the fact that from some $N$ $y_n ge c > 0$. This lemma is actually one of the several of the same kind in the exercise before limit.
$endgroup$
– roman
Dec 6 '18 at 19:32
$begingroup$
I may be mistaken but I think you need more constraints on $c$. It seems like there's nothing stopping us from having $c=1/varepsilon>0$, in which case $x_ny_n>cvarepsilon=1$ doesn't tell us much.
$endgroup$
– Jam
Dec 6 '18 at 19:39
$begingroup$
In other words, you've shown that $x$ is unbounded above and $y$ is bounded from below but how do we know that all $y$ have the same lower bound? And how do we know that the lower bound of $y$ doesn't get small very quickly (counteracting the size of $varepsilon$)?
$endgroup$
– Jam
Dec 6 '18 at 19:42
$begingroup$
@Jam, I have updated the question and added more context of where that lemma came from
$endgroup$
– roman
Dec 6 '18 at 22:10
|
show 2 more comments
$begingroup$
What's $c$ in your lemma? It feels like it's just randomly introduced?
$endgroup$
– Jam
Dec 6 '18 at 19:30
$begingroup$
@Jam, the exercise to prove this lemma comes right before the exercise on the limit (which in in the question section). No constraints for $c$ are given except for the fact that from some $N$ $y_n ge c > 0$. This lemma is actually one of the several of the same kind in the exercise before limit.
$endgroup$
– roman
Dec 6 '18 at 19:32
$begingroup$
I may be mistaken but I think you need more constraints on $c$. It seems like there's nothing stopping us from having $c=1/varepsilon>0$, in which case $x_ny_n>cvarepsilon=1$ doesn't tell us much.
$endgroup$
– Jam
Dec 6 '18 at 19:39
$begingroup$
In other words, you've shown that $x$ is unbounded above and $y$ is bounded from below but how do we know that all $y$ have the same lower bound? And how do we know that the lower bound of $y$ doesn't get small very quickly (counteracting the size of $varepsilon$)?
$endgroup$
– Jam
Dec 6 '18 at 19:42
$begingroup$
@Jam, I have updated the question and added more context of where that lemma came from
$endgroup$
– roman
Dec 6 '18 at 22:10
$begingroup$
What's $c$ in your lemma? It feels like it's just randomly introduced?
$endgroup$
– Jam
Dec 6 '18 at 19:30
$begingroup$
What's $c$ in your lemma? It feels like it's just randomly introduced?
$endgroup$
– Jam
Dec 6 '18 at 19:30
$begingroup$
@Jam, the exercise to prove this lemma comes right before the exercise on the limit (which in in the question section). No constraints for $c$ are given except for the fact that from some $N$ $y_n ge c > 0$. This lemma is actually one of the several of the same kind in the exercise before limit.
$endgroup$
– roman
Dec 6 '18 at 19:32
$begingroup$
@Jam, the exercise to prove this lemma comes right before the exercise on the limit (which in in the question section). No constraints for $c$ are given except for the fact that from some $N$ $y_n ge c > 0$. This lemma is actually one of the several of the same kind in the exercise before limit.
$endgroup$
– roman
Dec 6 '18 at 19:32
$begingroup$
I may be mistaken but I think you need more constraints on $c$. It seems like there's nothing stopping us from having $c=1/varepsilon>0$, in which case $x_ny_n>cvarepsilon=1$ doesn't tell us much.
$endgroup$
– Jam
Dec 6 '18 at 19:39
$begingroup$
I may be mistaken but I think you need more constraints on $c$. It seems like there's nothing stopping us from having $c=1/varepsilon>0$, in which case $x_ny_n>cvarepsilon=1$ doesn't tell us much.
$endgroup$
– Jam
Dec 6 '18 at 19:39
$begingroup$
In other words, you've shown that $x$ is unbounded above and $y$ is bounded from below but how do we know that all $y$ have the same lower bound? And how do we know that the lower bound of $y$ doesn't get small very quickly (counteracting the size of $varepsilon$)?
$endgroup$
– Jam
Dec 6 '18 at 19:42
$begingroup$
In other words, you've shown that $x$ is unbounded above and $y$ is bounded from below but how do we know that all $y$ have the same lower bound? And how do we know that the lower bound of $y$ doesn't get small very quickly (counteracting the size of $varepsilon$)?
$endgroup$
– Jam
Dec 6 '18 at 19:42
$begingroup$
@Jam, I have updated the question and added more context of where that lemma came from
$endgroup$
– roman
Dec 6 '18 at 22:10
$begingroup$
@Jam, I have updated the question and added more context of where that lemma came from
$endgroup$
– roman
Dec 6 '18 at 22:10
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
All you need here is:
$sqrt{n^2-1}ge n-1$ if $nge 1$ (to see this, just square both sides);
$sqrt nle n/2$ if $nge 4$ (to see this, just square both sides).
So $sqrt{n^2-1}-sqrt n ge n/2-1$ if $nge 4$.
And $lim_{nrightarrowinfty}(n/2-1) = +infty$.
answered Dec 6 '18 at 22:39
TonyKTonyK
42.5k355134
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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$begingroup$
All you need here is:
$sqrt{n^2-1}ge n-1$ if $nge 1$ (to see this, just square both sides);
$sqrt nle n/2$ if $nge 4$ (to see this, just square both sides).
So $sqrt{n^2-1}-sqrt n ge n/2-1$ if $nge 4$.
And $lim_{nrightarrowinfty}(n/2-1) = +infty$.
answered Dec 6 '18 at 22:39
TonyKTonyK
42.5k355134
$endgroup$
add a comment |
$begingroup$
All you need here is:
$sqrt{n^2-1}ge n-1$ if $nge 1$ (to see this, just square both sides);
$sqrt nle n/2$ if $nge 4$ (to see this, just square both sides).
So $sqrt{n^2-1}-sqrt n ge n/2-1$ if $nge 4$.
And $lim_{nrightarrowinfty}(n/2-1) = +infty$.
answered Dec 6 '18 at 22:39
TonyKTonyK
42.5k355134
$endgroup$
add a comment |
$begingroup$
All you need here is:
$sqrt{n^2-1}ge n-1$ if $nge 1$ (to see this, just square both sides);
$sqrt nle n/2$ if $nge 4$ (to see this, just square both sides).
So $sqrt{n^2-1}-sqrt n ge n/2-1$ if $nge 4$.
And $lim_{nrightarrowinfty}(n/2-1) = +infty$.
answered Dec 6 '18 at 22:39
TonyKTonyK
42.5k355134
$endgroup$
All you need here is:
$sqrt{n^2-1}ge n-1$ if $nge 1$ (to see this, just square both sides);
$sqrt nle n/2$ if $nge 4$ (to see this, just square both sides).
So $sqrt{n^2-1}-sqrt n ge n/2-1$ if $nge 4$.
And $lim_{nrightarrowinfty}(n/2-1) = +infty$.
answered Dec 6 '18 at 22:39
TonyKTonyK
42.5k355134
answered Dec 6 '18 at 22:39
TonyKTonyK
42.5k355134
answered Dec 6 '18 at 22:39
TonyKTonyK
42.5k355134
answered Dec 6 '18 at 22:39
TonyKTonyK
42.5k355134
42.5k355134
add a comment |
add a comment |
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$begingroup$
What's $c$ in your lemma? It feels like it's just randomly introduced?
$endgroup$
– Jam
Dec 6 '18 at 19:30
$begingroup$
@Jam, the exercise to prove this lemma comes right before the exercise on the limit (which in in the question section). No constraints for $c$ are given except for the fact that from some $N$ $y_n ge c > 0$. This lemma is actually one of the several of the same kind in the exercise before limit.
$endgroup$
– roman
Dec 6 '18 at 19:32
$begingroup$
I may be mistaken but I think you need more constraints on $c$. It seems like there's nothing stopping us from having $c=1/varepsilon>0$, in which case $x_ny_n>cvarepsilon=1$ doesn't tell us much.
$endgroup$
– Jam
Dec 6 '18 at 19:39
$begingroup$
In other words, you've shown that $x$ is unbounded above and $y$ is bounded from below but how do we know that all $y$ have the same lower bound? And how do we know that the lower bound of $y$ doesn't get small very quickly (counteracting the size of $varepsilon$)?
$endgroup$
– Jam
Dec 6 '18 at 19:42
$begingroup$
@Jam, I have updated the question and added more context of where that lemma came from
$endgroup$
– roman
Dec 6 '18 at 22:10