Is this way of finding $limlimits_{xto +infty}(x-ln(x^2+1))$ valid?












2












$begingroup$


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):enter image description here



If it's not valid, what would be a valid way to find the limit?






real-analysis calculus limits continuity







share|cite|improve this question











$endgroup$








  • 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
















2












$begingroup$


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):enter image description here



If it's not valid, what would be a valid way to find the limit?






real-analysis calculus limits continuity







share|cite|improve this question











$endgroup$








  • 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














2












2








2





$begingroup$


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):enter image description here



If it's not valid, what would be a valid way to find the limit?






real-analysis calculus limits continuity







share|cite|improve this question











$endgroup$




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):enter image description here



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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 23:03









rtybase

11.5k31534




11.5k31534










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














  • 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










2 Answers
2






active

oldest

votes


















1












$begingroup$

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 $xto infty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















1












$begingroup$

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 $x=e^{C}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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 $xto infty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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 $xto infty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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 $xto infty$.






share|cite|improve this answer









$endgroup$



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 $xto infty$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 5 at 22:52









Foobaz JohnFoobaz John

22.9k41552




22.9k41552












  • $begingroup$
    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




$begingroup$
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











1












$begingroup$

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 $x=e^{C}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















1












$begingroup$

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 $x=e^{C}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












1








1





$begingroup$

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 $x=e^{C}$.






share|cite|improve this answer









$endgroup$



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 $x=e^{C}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 5 at 23:28









Kavi Rama MurthyKavi Rama Murthy

72.9k53170




72.9k53170












  • $begingroup$
    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










  • $begingroup$
    @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




















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