Is this way of finding $limlimits_{xto +infty}(x-ln(x^2+1))$ valid?
$begingroup$
I needed to find: $$limlimits_{xto +infty}(x-ln(x^2+1))$$
So here are the steps I took:
Step 1: Replace $x$ with $ln(e^x)$:
$$limlimits_{xto +infty}left(ln(e^x)-ln(x^2+1)right)$$
$$limlimits_{xto +infty}lnleft(frac{e^x}{x^2+1}right)$$
Step 2: Bring the limit inside of the natural log function since it is continuous on the required interval.
$$lnleft(lim_{xto +infty}frac{e^x}{x^2+1}right)$$
Step 3: Apply L'Hospital's rule twice and evaluate:
$$lnleft(lim_{xto +infty}e^xright)$$
$$ln(+infty) = +infty$$
My question is whether step 2 is valid here because $limlimits_{xto infty}frac{e^x}{x^2 + 1}$ doesn't exist (its $+infty$), and in order to move the limit operator inside the function the limit $limlimits_{xto infty}frac{e^x}{x^2 + 1}$ must exist according to this theorem in a book about Calculus (ISBN 978-0-470-64769-1):
If it's not valid, what would be a valid way to find the limit?
real-analysis calculus limits continuity
edited Jan 5 at 23:03
rtybase
11.5k31534
asked Jan 5 at 22:40
user3071028user3071028
826
$endgroup$
add a comment |
$begingroup$
I needed to find: $$limlimits_{xto +infty}(x-ln(x^2+1))$$
So here are the steps I took:
Step 1: Replace $x$ with $ln(e^x)$:
$$limlimits_{xto +infty}left(ln(e^x)-ln(x^2+1)right)$$
$$limlimits_{xto +infty}lnleft(frac{e^x}{x^2+1}right)$$
Step 2: Bring the limit inside of the natural log function since it is continuous on the required interval.
$$lnleft(lim_{xto +infty}frac{e^x}{x^2+1}right)$$
Step 3: Apply L'Hospital's rule twice and evaluate:
$$lnleft(lim_{xto +infty}e^xright)$$
$$ln(+infty) = +infty$$
My question is whether step 2 is valid here because $limlimits_{xto infty}frac{e^x}{x^2 + 1}$ doesn't exist (its $+infty$), and in order to move the limit operator inside the function the limit $limlimits_{xto infty}frac{e^x}{x^2 + 1}$ must exist according to this theorem in a book about Calculus (ISBN 978-0-470-64769-1):
If it's not valid, what would be a valid way to find the limit?
real-analysis calculus limits continuity
edited Jan 5 at 23:03
rtybase
11.5k31534
asked Jan 5 at 22:40
user3071028user3071028
826
$endgroup$
1
$begingroup$
Your step is valid. The book deals with the theorem where the limit does exist. Similar result holds if the limit does not exist. See the theorem mentioned at the end of this answer: math.stackexchange.com/a/1073047/72031
$endgroup$
– Paramanand Singh
Jan 6 at 2:24
add a comment |
$begingroup$
I needed to find: $$limlimits_{xto +infty}(x-ln(x^2+1))$$
So here are the steps I took:
Step 1: Replace $x$ with $ln(e^x)$:
$$limlimits_{xto +infty}left(ln(e^x)-ln(x^2+1)right)$$
$$limlimits_{xto +infty}lnleft(frac{e^x}{x^2+1}right)$$
Step 2: Bring the limit inside of the natural log function since it is continuous on the required interval.
$$lnleft(lim_{xto +infty}frac{e^x}{x^2+1}right)$$
Step 3: Apply L'Hospital's rule twice and evaluate:
$$lnleft(lim_{xto +infty}e^xright)$$
$$ln(+infty) = +infty$$
My question is whether step 2 is valid here because $limlimits_{xto infty}frac{e^x}{x^2 + 1}$ doesn't exist (its $+infty$), and in order to move the limit operator inside the function the limit $limlimits_{xto infty}frac{e^x}{x^2 + 1}$ must exist according to this theorem in a book about Calculus (ISBN 978-0-470-64769-1):
If it's not valid, what would be a valid way to find the limit?
real-analysis calculus limits continuity
edited Jan 5 at 23:03
rtybase
11.5k31534
asked Jan 5 at 22:40
user3071028user3071028
826
$endgroup$
I needed to find: $$limlimits_{xto +infty}(x-ln(x^2+1))$$
So here are the steps I took:
Step 1: Replace $x$ with $ln(e^x)$:
$$limlimits_{xto +infty}left(ln(e^x)-ln(x^2+1)right)$$
$$limlimits_{xto +infty}lnleft(frac{e^x}{x^2+1}right)$$
Step 2: Bring the limit inside of the natural log function since it is continuous on the required interval.
$$lnleft(lim_{xto +infty}frac{e^x}{x^2+1}right)$$
Step 3: Apply L'Hospital's rule twice and evaluate:
$$lnleft(lim_{xto +infty}e^xright)$$
$$ln(+infty) = +infty$$
My question is whether step 2 is valid here because $limlimits_{xto infty}frac{e^x}{x^2 + 1}$ doesn't exist (its $+infty$), and in order to move the limit operator inside the function the limit $limlimits_{xto infty}frac{e^x}{x^2 + 1}$ must exist according to this theorem in a book about Calculus (ISBN 978-0-470-64769-1):
If it's not valid, what would be a valid way to find the limit?
real-analysis calculus limits continuity
real-analysis calculus limits continuity
edited Jan 5 at 23:03
rtybase
11.5k31534
asked Jan 5 at 22:40
user3071028user3071028
826
edited Jan 5 at 23:03
rtybase
11.5k31534
asked Jan 5 at 22:40
user3071028user3071028
826
edited Jan 5 at 23:03
rtybase
11.5k31534
edited Jan 5 at 23:03
rtybase
11.5k31534
edited Jan 5 at 23:03
rtybase
11.5k31534
11.5k31534
asked Jan 5 at 22:40
user3071028user3071028
826
asked Jan 5 at 22:40
user3071028user3071028
826
asked Jan 5 at 22:40
user3071028user3071028
826
826
1
$begingroup$
Your step is valid. The book deals with the theorem where the limit does exist. Similar result holds if the limit does not exist. See the theorem mentioned at the end of this answer: math.stackexchange.com/a/1073047/72031
$endgroup$
– Paramanand Singh
Jan 6 at 2:24
add a comment |
1
$begingroup$
Your step is valid. The book deals with the theorem where the limit does exist. Similar result holds if the limit does not exist. See the theorem mentioned at the end of this answer: math.stackexchange.com/a/1073047/72031
$endgroup$
– Paramanand Singh
Jan 6 at 2:24
1
1
$begingroup$
Your step is valid. The book deals with the theorem where the limit does exist. Similar result holds if the limit does not exist. See the theorem mentioned at the end of this answer: math.stackexchange.com/a/1073047/72031
$endgroup$
– Paramanand Singh
Jan 6 at 2:24
$begingroup$
Your step is valid. The book deals with the theorem where the limit does exist. Similar result holds if the limit does not exist. See the theorem mentioned at the end of this answer: math.stackexchange.com/a/1073047/72031
$endgroup$
– Paramanand Singh
Jan 6 at 2:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $M>0$ be given. Since $ln(u)to infty$ as $utoinfty$, there exists $K$ such that
for all $x$ with $frac{e^x}{x^2+1}>K$ it follows that $lnleft(frac{e^x}{x^2+1}right)>M$. Since $frac{e^x}{x^2+1}to infty$ as $xto infty$, there exists $N$ such that
$$
x>Nimplies frac{e^x}{x^2+1}>Kimplies lnleft(frac{e^x}{x^2+1}right)>M.
$$
By definition of a limit it follows that
$$
lim_{xtoinfty}frac{e^x}{x^2+1}=infty.
$$
Note we can mimic the same argument to conclude that if $f(x)to infty$ as $xto infty$ and $g(x)to infty$ as $xto infty$, then $g(f(x))to infty$ as $xto infty$.
answered Jan 5 at 22:52
Foobaz JohnFoobaz John
22.8k41452
$endgroup$
$begingroup$
That's a nice clean way, thanks! Can you please explicitly answer the first question: whether the way I did it is mathematically valid or invalid?
$endgroup$
– user3071028
Jan 6 at 0:19
add a comment |
$begingroup$
In step 2 all you need is the fact that $ln (y) to infty$ as $yto infty$. Since $frac {e^{x}} {x^{2}+1} to infty$ it follows that $ln (frac {e^{x}} {x^{2}+1}) to infty$. To prove that $ln (y) to infty$ as $yto infty$ assume that this is false. Since $ln , x$is an increasing function, if it doesn't not tend to infinity, it would be bounded for $x>1$, say $ln, x <C$ for all $x>1$. You get a contradiction from this if you take $x=e^{C}$.
answered Jan 5 at 23:28
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
$endgroup$
$begingroup$
Thank you! Can you please explicitly answer the first question: whether the way I did it is mathematically valid or invalid?
$endgroup$
– user3071028
Jan 6 at 0:22
$begingroup$
@user3071028 Everything you have done is right if you use that fact that $ln , yto infty$ as $ y to infty$.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 0:35
add a comment |
Your Answer
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
Let $M>0$ be given. Since $ln(u)to infty$ as $utoinfty$, there exists $K$ such that
for all $x$ with $frac{e^x}{x^2+1}>K$ it follows that $lnleft(frac{e^x}{x^2+1}right)>M$. Since $frac{e^x}{x^2+1}to infty$ as $xto infty$, there exists $N$ such that
$$
x>Nimplies frac{e^x}{x^2+1}>Kimplies lnleft(frac{e^x}{x^2+1}right)>M.
$$
By definition of a limit it follows that
$$
lim_{xtoinfty}frac{e^x}{x^2+1}=infty.
$$
Note we can mimic the same argument to conclude that if $f(x)to infty$ as $xto infty$ and $g(x)to infty$ as $xto infty$, then $g(f(x))to infty$ as $xto infty$.
answered Jan 5 at 22:52
Foobaz JohnFoobaz John
22.8k41452
$endgroup$
$begingroup$
That's a nice clean way, thanks! Can you please explicitly answer the first question: whether the way I did it is mathematically valid or invalid?
$endgroup$
– user3071028
Jan 6 at 0:19
add a comment |
$begingroup$
Let $M>0$ be given. Since $ln(u)to infty$ as $utoinfty$, there exists $K$ such that
for all $x$ with $frac{e^x}{x^2+1}>K$ it follows that $lnleft(frac{e^x}{x^2+1}right)>M$. Since $frac{e^x}{x^2+1}to infty$ as $xto infty$, there exists $N$ such that
$$
x>Nimplies frac{e^x}{x^2+1}>Kimplies lnleft(frac{e^x}{x^2+1}right)>M.
$$
By definition of a limit it follows that
$$
lim_{xtoinfty}frac{e^x}{x^2+1}=infty.
$$
Note we can mimic the same argument to conclude that if $f(x)to infty$ as $xto infty$ and $g(x)to infty$ as $xto infty$, then $g(f(x))to infty$ as $xto infty$.
answered Jan 5 at 22:52
Foobaz JohnFoobaz John
22.8k41452
$endgroup$
$begingroup$
That's a nice clean way, thanks! Can you please explicitly answer the first question: whether the way I did it is mathematically valid or invalid?
$endgroup$
– user3071028
Jan 6 at 0:19
add a comment |
$begingroup$
Let $M>0$ be given. Since $ln(u)to infty$ as $utoinfty$, there exists $K$ such that
for all $x$ with $frac{e^x}{x^2+1}>K$ it follows that $lnleft(frac{e^x}{x^2+1}right)>M$. Since $frac{e^x}{x^2+1}to infty$ as $xto infty$, there exists $N$ such that
$$
x>Nimplies frac{e^x}{x^2+1}>Kimplies lnleft(frac{e^x}{x^2+1}right)>M.
$$
By definition of a limit it follows that
$$
lim_{xtoinfty}frac{e^x}{x^2+1}=infty.
$$
Note we can mimic the same argument to conclude that if $f(x)to infty$ as $xto infty$ and $g(x)to infty$ as $xto infty$, then $g(f(x))to infty$ as $xto infty$.
answered Jan 5 at 22:52
Foobaz JohnFoobaz John
22.8k41452
$endgroup$
Let $M>0$ be given. Since $ln(u)to infty$ as $utoinfty$, there exists $K$ such that
for all $x$ with $frac{e^x}{x^2+1}>K$ it follows that $lnleft(frac{e^x}{x^2+1}right)>M$. Since $frac{e^x}{x^2+1}to infty$ as $xto infty$, there exists $N$ such that
$$
x>Nimplies frac{e^x}{x^2+1}>Kimplies lnleft(frac{e^x}{x^2+1}right)>M.
$$
By definition of a limit it follows that
$$
lim_{xtoinfty}frac{e^x}{x^2+1}=infty.
$$
Note we can mimic the same argument to conclude that if $f(x)to infty$ as $xto infty$ and $g(x)to infty$ as $xto infty$, then $g(f(x))to infty$ as $xto infty$.
answered Jan 5 at 22:52
Foobaz JohnFoobaz John
22.8k41452
answered Jan 5 at 22:52
Foobaz JohnFoobaz John
22.8k41452
answered Jan 5 at 22:52
Foobaz JohnFoobaz John
22.8k41452
answered Jan 5 at 22:52
Foobaz JohnFoobaz John
22.8k41452
22.8k41452
$begingroup$
That's a nice clean way, thanks! Can you please explicitly answer the first question: whether the way I did it is mathematically valid or invalid?
$endgroup$
– user3071028
Jan 6 at 0:19
add a comment |
$begingroup$
That's a nice clean way, thanks! Can you please explicitly answer the first question: whether the way I did it is mathematically valid or invalid?
$endgroup$
– user3071028
Jan 6 at 0:19
$begingroup$
That's a nice clean way, thanks! Can you please explicitly answer the first question: whether the way I did it is mathematically valid or invalid?
$endgroup$
– user3071028
Jan 6 at 0:19
$begingroup$
That's a nice clean way, thanks! Can you please explicitly answer the first question: whether the way I did it is mathematically valid or invalid?
$endgroup$
– user3071028
Jan 6 at 0:19
add a comment |
$begingroup$
In step 2 all you need is the fact that $ln (y) to infty$ as $yto infty$. Since $frac {e^{x}} {x^{2}+1} to infty$ it follows that $ln (frac {e^{x}} {x^{2}+1}) to infty$. To prove that $ln (y) to infty$ as $yto infty$ assume that this is false. Since $ln , x$is an increasing function, if it doesn't not tend to infinity, it would be bounded for $x>1$, say $ln, x <C$ for all $x>1$. You get a contradiction from this if you take $x=e^{C}$.
answered Jan 5 at 23:28
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
$endgroup$
$begingroup$
Thank you! Can you please explicitly answer the first question: whether the way I did it is mathematically valid or invalid?
$endgroup$
– user3071028
Jan 6 at 0:22
$begingroup$
@user3071028 Everything you have done is right if you use that fact that $ln , yto infty$ as $ y to infty$.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 0:35
add a comment |
$begingroup$
In step 2 all you need is the fact that $ln (y) to infty$ as $yto infty$. Since $frac {e^{x}} {x^{2}+1} to infty$ it follows that $ln (frac {e^{x}} {x^{2}+1}) to infty$. To prove that $ln (y) to infty$ as $yto infty$ assume that this is false. Since $ln , x$is an increasing function, if it doesn't not tend to infinity, it would be bounded for $x>1$, say $ln, x <C$ for all $x>1$. You get a contradiction from this if you take $x=e^{C}$.
answered Jan 5 at 23:28
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
$endgroup$
$begingroup$
Thank you! Can you please explicitly answer the first question: whether the way I did it is mathematically valid or invalid?
$endgroup$
– user3071028
Jan 6 at 0:22
$begingroup$
@user3071028 Everything you have done is right if you use that fact that $ln , yto infty$ as $ y to infty$.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 0:35
add a comment |
$begingroup$
In step 2 all you need is the fact that $ln (y) to infty$ as $yto infty$. Since $frac {e^{x}} {x^{2}+1} to infty$ it follows that $ln (frac {e^{x}} {x^{2}+1}) to infty$. To prove that $ln (y) to infty$ as $yto infty$ assume that this is false. Since $ln , x$is an increasing function, if it doesn't not tend to infinity, it would be bounded for $x>1$, say $ln, x <C$ for all $x>1$. You get a contradiction from this if you take $x=e^{C}$.
answered Jan 5 at 23:28
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
$endgroup$
In step 2 all you need is the fact that $ln (y) to infty$ as $yto infty$. Since $frac {e^{x}} {x^{2}+1} to infty$ it follows that $ln (frac {e^{x}} {x^{2}+1}) to infty$. To prove that $ln (y) to infty$ as $yto infty$ assume that this is false. Since $ln , x$is an increasing function, if it doesn't not tend to infinity, it would be bounded for $x>1$, say $ln, x <C$ for all $x>1$. You get a contradiction from this if you take $x=e^{C}$.
answered Jan 5 at 23:28
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
answered Jan 5 at 23:28
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
answered Jan 5 at 23:28
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
answered Jan 5 at 23:28
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
68.6k53169
$begingroup$
Thank you! Can you please explicitly answer the first question: whether the way I did it is mathematically valid or invalid?
$endgroup$
– user3071028
Jan 6 at 0:22
$begingroup$
@user3071028 Everything you have done is right if you use that fact that $ln , yto infty$ as $ y to infty$.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 0:35
add a comment |
$begingroup$
Thank you! Can you please explicitly answer the first question: whether the way I did it is mathematically valid or invalid?
$endgroup$
– user3071028
Jan 6 at 0:22
$begingroup$
@user3071028 Everything you have done is right if you use that fact that $ln , yto infty$ as $ y to infty$.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 0:35
$begingroup$
Thank you! Can you please explicitly answer the first question: whether the way I did it is mathematically valid or invalid?
$endgroup$
– user3071028
Jan 6 at 0:22
$begingroup$
Thank you! Can you please explicitly answer the first question: whether the way I did it is mathematically valid or invalid?
$endgroup$
– user3071028
Jan 6 at 0:22
$begingroup$
@user3071028 Everything you have done is right if you use that fact that $ln , yto infty$ as $ y to infty$.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 0:35
$begingroup$
@user3071028 Everything you have done is right if you use that fact that $ln , yto infty$ as $ y to infty$.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 0:35
add a comment |
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$begingroup$
Your step is valid. The book deals with the theorem where the limit does exist. Similar result holds if the limit does not exist. See the theorem mentioned at the end of this answer: math.stackexchange.com/a/1073047/72031
$endgroup$
– Paramanand Singh
Jan 6 at 2:24