Mixing
determine if the statement with limit is true
$begingroup$
I want to determine if the statement
$$lim_{xto a} f(x) = infty Rightarrow lim_{xto a}frac{1}{f(x)} = 0$$
is true or not (by proving it or proving a contradiction).
I know that I have a definition of limits that could be of help.
That is, if the function has a limit $A$ when $x rightarrow a$ then there is a number $ε > 0$ and $w$ such that
$$|x-a|< w Rightarrow |f(x)-A| < ε . $$
And I have all the other rules for limits. But I don't know how to go about this problem.
limits analysis proof-writing
edited Jan 23 at 22:02
J. W. Tanner
3,2401320
asked Jan 23 at 19:02
F WiF Wi
384
$endgroup$
add a comment |
$begingroup$
I want to determine if the statement
$$lim_{xto a} f(x) = infty Rightarrow lim_{xto a}frac{1}{f(x)} = 0$$
is true or not (by proving it or proving a contradiction).
I know that I have a definition of limits that could be of help.
That is, if the function has a limit $A$ when $x rightarrow a$ then there is a number $ε > 0$ and $w$ such that
$$|x-a|< w Rightarrow |f(x)-A| < ε . $$
And I have all the other rules for limits. But I don't know how to go about this problem.
limits analysis proof-writing
edited Jan 23 at 22:02
J. W. Tanner
3,2401320
asked Jan 23 at 19:02
F WiF Wi
384
$endgroup$
add a comment |
$begingroup$
I want to determine if the statement
$$lim_{xto a} f(x) = infty Rightarrow lim_{xto a}frac{1}{f(x)} = 0$$
is true or not (by proving it or proving a contradiction).
I know that I have a definition of limits that could be of help.
That is, if the function has a limit $A$ when $x rightarrow a$ then there is a number $ε > 0$ and $w$ such that
$$|x-a|< w Rightarrow |f(x)-A| < ε . $$
And I have all the other rules for limits. But I don't know how to go about this problem.
limits analysis proof-writing
edited Jan 23 at 22:02
J. W. Tanner
3,2401320
asked Jan 23 at 19:02
F WiF Wi
384
$endgroup$
I want to determine if the statement
$$lim_{xto a} f(x) = infty Rightarrow lim_{xto a}frac{1}{f(x)} = 0$$
is true or not (by proving it or proving a contradiction).
I know that I have a definition of limits that could be of help.
That is, if the function has a limit $A$ when $x rightarrow a$ then there is a number $ε > 0$ and $w$ such that
$$|x-a|< w Rightarrow |f(x)-A| < ε . $$
And I have all the other rules for limits. But I don't know how to go about this problem.
limits analysis proof-writing
limits analysis proof-writing
edited Jan 23 at 22:02
J. W. Tanner
3,2401320
asked Jan 23 at 19:02
F WiF Wi
384
edited Jan 23 at 22:02
J. W. Tanner
3,2401320
asked Jan 23 at 19:02
F WiF Wi
384
edited Jan 23 at 22:02
J. W. Tanner
3,2401320
edited Jan 23 at 22:02
J. W. Tanner
3,2401320
edited Jan 23 at 22:02
J. W. Tanner
3,2401320
3,2401320
asked Jan 23 at 19:02
F WiF Wi
384
asked Jan 23 at 19:02
F WiF Wi
384
asked Jan 23 at 19:02
F WiF Wi
384
384
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Alternatively we may use Heine's definition of the limit of a sequence.
Let $x_n$ be a sequence so that $x_n rightarrow a => lim_{nto infty} f(x_n)=infty$.
We have: $lim_{xto a} frac{1}{f(x)}=lim_{nto infty} frac{1}{f(x_n)}=frac{1}{lim_{ntoinfty} f(x_n)}=frac{1}{infty}=0$
answered Jan 24 at 12:30
AlexdanutAlexdanut
1438
$endgroup$
add a comment |
$begingroup$
HINT
We have to show that for each $epsilon > 0 exists delta > 0$ such that whenever $|x-a| < delta$ you have
$$
epsilon
> left|frac{1}{f(x)} - 0 right|
= left|frac1{f(x)}right|
iff |f(x)| > 1/epsilon.
$$
Assume that as $x to a$, we have $f(x) to infty$, in other words, that $forall N > 0 exists delta>0$ such that $f(x) > N$ whenever $|x - a| < delta$.
Can you see how to pick $epsilon = epsilon(N)$ to make what you want happen?
answered Jan 23 at 19:10
gt6989bgt6989b
34.9k22557
$endgroup$
$begingroup$
I know that when proving that a specific limit exists for a function you want to find δ that depends on ϵ so δ = δ(ϵ) but I cant seem to figure out why in this case you would want to find ϵ=ϵ(N)
$endgroup$
– F Wi
Jan 23 at 22:01
$begingroup$
@FWi knowing the $N$ (existence of which is assumed but value unknown) allows to construct the $epsilon$ with the needed property. It is perfectly fine that $epsilon$ is a function of $N$, which is in tern and function of $delta$...
$endgroup$
– gt6989b
Jan 24 at 4:38
add a comment |
$begingroup$
We say that $lim_{xto a} f(x) = infty$, if $forall M>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $f(x) > M $.
We say that $lim_{xto a}frac{1}{f(x)} = 0$, if $forall ε>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $left|frac{1}{f(x)} - 0 right| <epsilon$.
It suffices to choose $epsilon = frac{1}{M}$, in which case $left|frac{1}{f(x)} - 0 right| <epsilon$ holds.
answered Jan 23 at 19:43
AlexandrosAlexandros
9401412
$endgroup$
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
$begingroup$
Alternatively we may use Heine's definition of the limit of a sequence.
Let $x_n$ be a sequence so that $x_n rightarrow a => lim_{nto infty} f(x_n)=infty$.
We have: $lim_{xto a} frac{1}{f(x)}=lim_{nto infty} frac{1}{f(x_n)}=frac{1}{lim_{ntoinfty} f(x_n)}=frac{1}{infty}=0$
answered Jan 24 at 12:30
AlexdanutAlexdanut
1438
$endgroup$
add a comment |
$begingroup$
Alternatively we may use Heine's definition of the limit of a sequence.
Let $x_n$ be a sequence so that $x_n rightarrow a => lim_{nto infty} f(x_n)=infty$.
We have: $lim_{xto a} frac{1}{f(x)}=lim_{nto infty} frac{1}{f(x_n)}=frac{1}{lim_{ntoinfty} f(x_n)}=frac{1}{infty}=0$
answered Jan 24 at 12:30
AlexdanutAlexdanut
1438
$endgroup$
add a comment |
$begingroup$
Alternatively we may use Heine's definition of the limit of a sequence.
Let $x_n$ be a sequence so that $x_n rightarrow a => lim_{nto infty} f(x_n)=infty$.
We have: $lim_{xto a} frac{1}{f(x)}=lim_{nto infty} frac{1}{f(x_n)}=frac{1}{lim_{ntoinfty} f(x_n)}=frac{1}{infty}=0$
answered Jan 24 at 12:30
AlexdanutAlexdanut
1438
$endgroup$
Alternatively we may use Heine's definition of the limit of a sequence.
Let $x_n$ be a sequence so that $x_n rightarrow a => lim_{nto infty} f(x_n)=infty$.
We have: $lim_{xto a} frac{1}{f(x)}=lim_{nto infty} frac{1}{f(x_n)}=frac{1}{lim_{ntoinfty} f(x_n)}=frac{1}{infty}=0$
answered Jan 24 at 12:30
AlexdanutAlexdanut
1438
answered Jan 24 at 12:30
AlexdanutAlexdanut
1438
answered Jan 24 at 12:30
AlexdanutAlexdanut
1438
answered Jan 24 at 12:30
AlexdanutAlexdanut
1438
1438
add a comment |
add a comment |
$begingroup$
HINT
We have to show that for each $epsilon > 0 exists delta > 0$ such that whenever $|x-a| < delta$ you have
$$
epsilon
> left|frac{1}{f(x)} - 0 right|
= left|frac1{f(x)}right|
iff |f(x)| > 1/epsilon.
$$
Assume that as $x to a$, we have $f(x) to infty$, in other words, that $forall N > 0 exists delta>0$ such that $f(x) > N$ whenever $|x - a| < delta$.
Can you see how to pick $epsilon = epsilon(N)$ to make what you want happen?
answered Jan 23 at 19:10
gt6989bgt6989b
34.9k22557
$endgroup$
$begingroup$
I know that when proving that a specific limit exists for a function you want to find δ that depends on ϵ so δ = δ(ϵ) but I cant seem to figure out why in this case you would want to find ϵ=ϵ(N)
$endgroup$
– F Wi
Jan 23 at 22:01
$begingroup$
@FWi knowing the $N$ (existence of which is assumed but value unknown) allows to construct the $epsilon$ with the needed property. It is perfectly fine that $epsilon$ is a function of $N$, which is in tern and function of $delta$...
$endgroup$
– gt6989b
Jan 24 at 4:38
add a comment |
$begingroup$
HINT
We have to show that for each $epsilon > 0 exists delta > 0$ such that whenever $|x-a| < delta$ you have
$$
epsilon
> left|frac{1}{f(x)} - 0 right|
= left|frac1{f(x)}right|
iff |f(x)| > 1/epsilon.
$$
Assume that as $x to a$, we have $f(x) to infty$, in other words, that $forall N > 0 exists delta>0$ such that $f(x) > N$ whenever $|x - a| < delta$.
Can you see how to pick $epsilon = epsilon(N)$ to make what you want happen?
answered Jan 23 at 19:10
gt6989bgt6989b
34.9k22557
$endgroup$
$begingroup$
I know that when proving that a specific limit exists for a function you want to find δ that depends on ϵ so δ = δ(ϵ) but I cant seem to figure out why in this case you would want to find ϵ=ϵ(N)
$endgroup$
– F Wi
Jan 23 at 22:01
$begingroup$
@FWi knowing the $N$ (existence of which is assumed but value unknown) allows to construct the $epsilon$ with the needed property. It is perfectly fine that $epsilon$ is a function of $N$, which is in tern and function of $delta$...
$endgroup$
– gt6989b
Jan 24 at 4:38
add a comment |
$begingroup$
HINT
We have to show that for each $epsilon > 0 exists delta > 0$ such that whenever $|x-a| < delta$ you have
$$
epsilon
> left|frac{1}{f(x)} - 0 right|
= left|frac1{f(x)}right|
iff |f(x)| > 1/epsilon.
$$
Assume that as $x to a$, we have $f(x) to infty$, in other words, that $forall N > 0 exists delta>0$ such that $f(x) > N$ whenever $|x - a| < delta$.
Can you see how to pick $epsilon = epsilon(N)$ to make what you want happen?
answered Jan 23 at 19:10
gt6989bgt6989b
34.9k22557
$endgroup$
HINT
We have to show that for each $epsilon > 0 exists delta > 0$ such that whenever $|x-a| < delta$ you have
$$
epsilon
> left|frac{1}{f(x)} - 0 right|
= left|frac1{f(x)}right|
iff |f(x)| > 1/epsilon.
$$
Assume that as $x to a$, we have $f(x) to infty$, in other words, that $forall N > 0 exists delta>0$ such that $f(x) > N$ whenever $|x - a| < delta$.
Can you see how to pick $epsilon = epsilon(N)$ to make what you want happen?
answered Jan 23 at 19:10
gt6989bgt6989b
34.9k22557
answered Jan 23 at 19:10
gt6989bgt6989b
34.9k22557
answered Jan 23 at 19:10
gt6989bgt6989b
34.9k22557
answered Jan 23 at 19:10
gt6989bgt6989b
34.9k22557
34.9k22557
$begingroup$
I know that when proving that a specific limit exists for a function you want to find δ that depends on ϵ so δ = δ(ϵ) but I cant seem to figure out why in this case you would want to find ϵ=ϵ(N)
$endgroup$
– F Wi
Jan 23 at 22:01
$begingroup$
@FWi knowing the $N$ (existence of which is assumed but value unknown) allows to construct the $epsilon$ with the needed property. It is perfectly fine that $epsilon$ is a function of $N$, which is in tern and function of $delta$...
$endgroup$
– gt6989b
Jan 24 at 4:38
add a comment |
$begingroup$
I know that when proving that a specific limit exists for a function you want to find δ that depends on ϵ so δ = δ(ϵ) but I cant seem to figure out why in this case you would want to find ϵ=ϵ(N)
$endgroup$
– F Wi
Jan 23 at 22:01
$begingroup$
@FWi knowing the $N$ (existence of which is assumed but value unknown) allows to construct the $epsilon$ with the needed property. It is perfectly fine that $epsilon$ is a function of $N$, which is in tern and function of $delta$...
$endgroup$
– gt6989b
Jan 24 at 4:38
$begingroup$
I know that when proving that a specific limit exists for a function you want to find δ that depends on ϵ so δ = δ(ϵ) but I cant seem to figure out why in this case you would want to find ϵ=ϵ(N)
$endgroup$
– F Wi
Jan 23 at 22:01
$begingroup$
I know that when proving that a specific limit exists for a function you want to find δ that depends on ϵ so δ = δ(ϵ) but I cant seem to figure out why in this case you would want to find ϵ=ϵ(N)
$endgroup$
– F Wi
Jan 23 at 22:01
$begingroup$
@FWi knowing the $N$ (existence of which is assumed but value unknown) allows to construct the $epsilon$ with the needed property. It is perfectly fine that $epsilon$ is a function of $N$, which is in tern and function of $delta$...
$endgroup$
– gt6989b
Jan 24 at 4:38
$begingroup$
@FWi knowing the $N$ (existence of which is assumed but value unknown) allows to construct the $epsilon$ with the needed property. It is perfectly fine that $epsilon$ is a function of $N$, which is in tern and function of $delta$...
$endgroup$
– gt6989b
Jan 24 at 4:38
add a comment |
$begingroup$
We say that $lim_{xto a} f(x) = infty$, if $forall M>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $f(x) > M $.
We say that $lim_{xto a}frac{1}{f(x)} = 0$, if $forall ε>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $left|frac{1}{f(x)} - 0 right| <epsilon$.
It suffices to choose $epsilon = frac{1}{M}$, in which case $left|frac{1}{f(x)} - 0 right| <epsilon$ holds.
answered Jan 23 at 19:43
AlexandrosAlexandros
9401412
$endgroup$
add a comment |
$begingroup$
We say that $lim_{xto a} f(x) = infty$, if $forall M>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $f(x) > M $.
We say that $lim_{xto a}frac{1}{f(x)} = 0$, if $forall ε>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $left|frac{1}{f(x)} - 0 right| <epsilon$.
It suffices to choose $epsilon = frac{1}{M}$, in which case $left|frac{1}{f(x)} - 0 right| <epsilon$ holds.
answered Jan 23 at 19:43
AlexandrosAlexandros
9401412
$endgroup$
add a comment |
$begingroup$
We say that $lim_{xto a} f(x) = infty$, if $forall M>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $f(x) > M $.
We say that $lim_{xto a}frac{1}{f(x)} = 0$, if $forall ε>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $left|frac{1}{f(x)} - 0 right| <epsilon$.
It suffices to choose $epsilon = frac{1}{M}$, in which case $left|frac{1}{f(x)} - 0 right| <epsilon$ holds.
answered Jan 23 at 19:43
AlexandrosAlexandros
9401412
$endgroup$
We say that $lim_{xto a} f(x) = infty$, if $forall M>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $f(x) > M $.
We say that $lim_{xto a}frac{1}{f(x)} = 0$, if $forall ε>0, exists δ>0$ s.t. whenever $vert x-a vert < δ$, then $left|frac{1}{f(x)} - 0 right| <epsilon$.
It suffices to choose $epsilon = frac{1}{M}$, in which case $left|frac{1}{f(x)} - 0 right| <epsilon$ holds.
answered Jan 23 at 19:43
AlexandrosAlexandros
9401412
answered Jan 23 at 19:43
AlexandrosAlexandros
9401412
answered Jan 23 at 19:43
AlexandrosAlexandros
9401412
answered Jan 23 at 19:43
AlexandrosAlexandros
9401412
9401412
add a comment |
add a comment |
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