Delta Epsilon Proof of $lim_{xtoinfty}frac{ln{x}^2}{x}=0$












6












$begingroup$


I am trying to prove that $lim_{xtoinfty}frac{ln{x}^2}{x}=0$ via the delta epsilon definition of a limit but I keep getting hung up.



I have started the proof like this:



$forall epsilon>0, exists delta$ so if $delta>0$, $|frac{ln{x}^2}{x}-0|<epsilon$.



If we suppose that $|frac{ln{x}^2}{x}-0|<epsilon$ and solve for x we get...



When I try solving the inequality for $x$, it keeps cancelling.



What I have tried:



$2x^{-1}ln{x}<ln{epsilon}$



Which becomes:



$2x^{-1}(x^1)<e^{ln{epsilon}}$



But then, $x^{-1}$ and $x^1$ multiply to $x^0$, which is $1$ and $x$ is removed from the problem.



Any pointers as how to proceed? I know that the limit is $0$, I just can't seem to get past solving for $x$.



Thank you.






calculus proof-verification epsilon-delta







share|cite|improve this question











$endgroup$












  • $begingroup$
    To typeset $x^{-1}$, use x^{-1}.
    $endgroup$
    – angryavian
    Dec 29 '18 at 2:32










  • $begingroup$
    Thank you @angryavian
    $endgroup$
    – limitsandlogs224
    Dec 29 '18 at 2:34










  • $begingroup$
    Hint: $ln (x^2)=2ln x$ for all $x>0.$
    $endgroup$
    – DanielWainfleet
    Dec 29 '18 at 6:36
















6












$begingroup$


I am trying to prove that $lim_{xtoinfty}frac{ln{x}^2}{x}=0$ via the delta epsilon definition of a limit but I keep getting hung up.



I have started the proof like this:



$forall epsilon>0, exists delta$ so if $delta>0$, $|frac{ln{x}^2}{x}-0|<epsilon$.



If we suppose that $|frac{ln{x}^2}{x}-0|<epsilon$ and solve for x we get...



When I try solving the inequality for $x$, it keeps cancelling.



What I have tried:



$2x^{-1}ln{x}<ln{epsilon}$



Which becomes:



$2x^{-1}(x^1)<e^{ln{epsilon}}$



But then, $x^{-1}$ and $x^1$ multiply to $x^0$, which is $1$ and $x$ is removed from the problem.



Any pointers as how to proceed? I know that the limit is $0$, I just can't seem to get past solving for $x$.



Thank you.






calculus proof-verification epsilon-delta







share|cite|improve this question











$endgroup$












  • $begingroup$
    To typeset $x^{-1}$, use x^{-1}.
    $endgroup$
    – angryavian
    Dec 29 '18 at 2:32










  • $begingroup$
    Thank you @angryavian
    $endgroup$
    – limitsandlogs224
    Dec 29 '18 at 2:34










  • $begingroup$
    Hint: $ln (x^2)=2ln x$ for all $x>0.$
    $endgroup$
    – DanielWainfleet
    Dec 29 '18 at 6:36














6












6








6





$begingroup$


I am trying to prove that $lim_{xtoinfty}frac{ln{x}^2}{x}=0$ via the delta epsilon definition of a limit but I keep getting hung up.



I have started the proof like this:



$forall epsilon>0, exists delta$ so if $delta>0$, $|frac{ln{x}^2}{x}-0|<epsilon$.



If we suppose that $|frac{ln{x}^2}{x}-0|<epsilon$ and solve for x we get...



When I try solving the inequality for $x$, it keeps cancelling.



What I have tried:



$2x^{-1}ln{x}<ln{epsilon}$



Which becomes:



$2x^{-1}(x^1)<e^{ln{epsilon}}$



But then, $x^{-1}$ and $x^1$ multiply to $x^0$, which is $1$ and $x$ is removed from the problem.



Any pointers as how to proceed? I know that the limit is $0$, I just can't seem to get past solving for $x$.



Thank you.






calculus proof-verification epsilon-delta







share|cite|improve this question











$endgroup$




I am trying to prove that $lim_{xtoinfty}frac{ln{x}^2}{x}=0$ via the delta epsilon definition of a limit but I keep getting hung up.



I have started the proof like this:



$forall epsilon>0, exists delta$ so if $delta>0$, $|frac{ln{x}^2}{x}-0|<epsilon$.



If we suppose that $|frac{ln{x}^2}{x}-0|<epsilon$ and solve for x we get...



When I try solving the inequality for $x$, it keeps cancelling.



What I have tried:



$2x^{-1}ln{x}<ln{epsilon}$



Which becomes:



$2x^{-1}(x^1)<e^{ln{epsilon}}$



But then, $x^{-1}$ and $x^1$ multiply to $x^0$, which is $1$ and $x$ is removed from the problem.



Any pointers as how to proceed? I know that the limit is $0$, I just can't seem to get past solving for $x$.



Thank you.






calculus proof-verification epsilon-delta




calculus proof-verification epsilon-delta






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 29 '18 at 5:09









The Count

2,31361431




2,31361431










asked Dec 29 '18 at 2:26









limitsandlogs224limitsandlogs224

576




576












  • $begingroup$
    To typeset $x^{-1}$, use x^{-1}.
    $endgroup$
    – angryavian
    Dec 29 '18 at 2:32










  • $begingroup$
    Thank you @angryavian
    $endgroup$
    – limitsandlogs224
    Dec 29 '18 at 2:34










  • $begingroup$
    Hint: $ln (x^2)=2ln x$ for all $x>0.$
    $endgroup$
    – DanielWainfleet
    Dec 29 '18 at 6:36


















  • $begingroup$
    To typeset $x^{-1}$, use x^{-1}.
    $endgroup$
    – angryavian
    Dec 29 '18 at 2:32










  • $begingroup$
    Thank you @angryavian
    $endgroup$
    – limitsandlogs224
    Dec 29 '18 at 2:34










  • $begingroup$
    Hint: $ln (x^2)=2ln x$ for all $x>0.$
    $endgroup$
    – DanielWainfleet
    Dec 29 '18 at 6:36
















$begingroup$
To typeset $x^{-1}$, use x^{-1}.
$endgroup$
– angryavian
Dec 29 '18 at 2:32




$begingroup$
To typeset $x^{-1}$, use x^{-1}.
$endgroup$
– angryavian
Dec 29 '18 at 2:32












$begingroup$
Thank you @angryavian
$endgroup$
– limitsandlogs224
Dec 29 '18 at 2:34




$begingroup$
Thank you @angryavian
$endgroup$
– limitsandlogs224
Dec 29 '18 at 2:34












$begingroup$
Hint: $ln (x^2)=2ln x$ for all $x>0.$
$endgroup$
– DanielWainfleet
Dec 29 '18 at 6:36




$begingroup$
Hint: $ln (x^2)=2ln x$ for all $x>0.$
$endgroup$
– DanielWainfleet
Dec 29 '18 at 6:36










2 Answers
2






active

oldest

votes


















2












$begingroup$


... I have started the proof like this:



$forall epsilon>0, exists delta$ so if $delta>0$, $|frac{ln{x}^2}{x}-0|<epsilon$.




The $delta$ in your attempt has nothing to do with $epsilon$.



By definition, given $epsilon>0$, you need to show that there exists $M>0$ such that
$$
x>M quad textbf{implies }quad |frac{2ln x}{x}|<epsilontag{1}
$$



Now for $x>1$,
$$
ln x=int_1^xfrac{1}{t} dtleq int_1^xfrac{1}{sqrt{t}} dt=2sqrt{t}big|_1^x
< 2sqrt{x},
$$
which implies that
$$
frac{2ln x}{x}<frac{4sqrt{x}}{x}=frac{4}{sqrt{x}},quad x>1.
$$
So you want
$
frac{4}{sqrt{x}}<epsilon,
$
which is true if
$$
x>(4/epsilon)^2.
$$
Now, setting
$M=(4/epsilon)^2$, you have the implication (1).






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    Clever way of showing $ln x<2sqrt x$ for $x>1$.
    $endgroup$
    – DanielWainfleet
    Dec 29 '18 at 6:39










  • $begingroup$
    What does $|_1^x$ mean?
    $endgroup$
    – limitsandlogs224
    Dec 31 '18 at 1:12










  • $begingroup$
    @limitsandlogs224: $f(x)|_a^b:=f(b)-f(a)$.
    $endgroup$
    – user587192
    Dec 31 '18 at 1:53



















4












$begingroup$

$ln x<2sqrt{x}$ so
$$left|dfrac{ln x^2}{x}right|<dfrac{4}{sqrt{x}}<dfrac{4}{sqrt{M}}leqvarepsilon$$
where $x>M$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How do you prove $ln x< 2sqrt{x}$?
    $endgroup$
    – user587192
    Dec 29 '18 at 2:52










  • $begingroup$
    $ln x<dfrac{x^alpha}{alpha}$ @user587192 with $alpha>0$.
    $endgroup$
    – Nosrati
    Dec 29 '18 at 3:01












  • $begingroup$
    Proof: $e^u>u$ therefore $u>ln u$ let $u=x^alpha$.
    $endgroup$
    – Nosrati
    Dec 29 '18 at 3:15










  • $begingroup$
    @Nosrati, thank you. Could you please explain how I could incorporate $frac{4}{x^{frac{1}{2}}}$ and $frac{4}{M^{frac{1}{2}}}$ into the proof? I thought that I had to solve $frac{4}{x{frac{1}{2}}} <epsilon$ for $x$ so any $epsilon$ is $>|frac{ln{x}^2}{x}-0|$ as long as $delta>frac{16}{epsilon^2}$. I see that they keep $x$ from getting eliminated but I just don't understand how those expressions get substituted into the inequality.
    $endgroup$
    – limitsandlogs224
    Dec 29 '18 at 3:45








  • 1




    $begingroup$
    In fact, it is not difficult to show that $log(x)<sqrt x$ for all $x>0$. Going one step further $log(x)le x^{1/e}$ for all $x$ with equality at $e^e$.
    $endgroup$
    – Mark Viola
    Dec 29 '18 at 20:20













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$


... I have started the proof like this:



$forall epsilon>0, exists delta$ so if $delta>0$, $|frac{ln{x}^2}{x}-0|<epsilon$.




The $delta$ in your attempt has nothing to do with $epsilon$.



By definition, given $epsilon>0$, you need to show that there exists $M>0$ such that
$$
x>M quad textbf{implies }quad |frac{2ln x}{x}|<epsilontag{1}
$$



Now for $x>1$,
$$
ln x=int_1^xfrac{1}{t} dtleq int_1^xfrac{1}{sqrt{t}} dt=2sqrt{t}big|_1^x
< 2sqrt{x},
$$
which implies that
$$
frac{2ln x}{x}<frac{4sqrt{x}}{x}=frac{4}{sqrt{x}},quad x>1.
$$
So you want
$
frac{4}{sqrt{x}}<epsilon,
$
which is true if
$$
x>(4/epsilon)^2.
$$
Now, setting
$M=(4/epsilon)^2$, you have the implication (1).






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    Clever way of showing $ln x<2sqrt x$ for $x>1$.
    $endgroup$
    – DanielWainfleet
    Dec 29 '18 at 6:39










  • $begingroup$
    What does $|_1^x$ mean?
    $endgroup$
    – limitsandlogs224
    Dec 31 '18 at 1:12










  • $begingroup$
    @limitsandlogs224: $f(x)|_a^b:=f(b)-f(a)$.
    $endgroup$
    – user587192
    Dec 31 '18 at 1:53
















2












$begingroup$


... I have started the proof like this:



$forall epsilon>0, exists delta$ so if $delta>0$, $|frac{ln{x}^2}{x}-0|<epsilon$.




The $delta$ in your attempt has nothing to do with $epsilon$.



By definition, given $epsilon>0$, you need to show that there exists $M>0$ such that
$$
x>M quad textbf{implies }quad |frac{2ln x}{x}|<epsilontag{1}
$$



Now for $x>1$,
$$
ln x=int_1^xfrac{1}{t} dtleq int_1^xfrac{1}{sqrt{t}} dt=2sqrt{t}big|_1^x
< 2sqrt{x},
$$
which implies that
$$
frac{2ln x}{x}<frac{4sqrt{x}}{x}=frac{4}{sqrt{x}},quad x>1.
$$
So you want
$
frac{4}{sqrt{x}}<epsilon,
$
which is true if
$$
x>(4/epsilon)^2.
$$
Now, setting
$M=(4/epsilon)^2$, you have the implication (1).






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    Clever way of showing $ln x<2sqrt x$ for $x>1$.
    $endgroup$
    – DanielWainfleet
    Dec 29 '18 at 6:39










  • $begingroup$
    What does $|_1^x$ mean?
    $endgroup$
    – limitsandlogs224
    Dec 31 '18 at 1:12










  • $begingroup$
    @limitsandlogs224: $f(x)|_a^b:=f(b)-f(a)$.
    $endgroup$
    – user587192
    Dec 31 '18 at 1:53














2












2








2





$begingroup$


... I have started the proof like this:



$forall epsilon>0, exists delta$ so if $delta>0$, $|frac{ln{x}^2}{x}-0|<epsilon$.




The $delta$ in your attempt has nothing to do with $epsilon$.



By definition, given $epsilon>0$, you need to show that there exists $M>0$ such that
$$
x>M quad textbf{implies }quad |frac{2ln x}{x}|<epsilontag{1}
$$



Now for $x>1$,
$$
ln x=int_1^xfrac{1}{t} dtleq int_1^xfrac{1}{sqrt{t}} dt=2sqrt{t}big|_1^x
< 2sqrt{x},
$$
which implies that
$$
frac{2ln x}{x}<frac{4sqrt{x}}{x}=frac{4}{sqrt{x}},quad x>1.
$$
So you want
$
frac{4}{sqrt{x}}<epsilon,
$
which is true if
$$
x>(4/epsilon)^2.
$$
Now, setting
$M=(4/epsilon)^2$, you have the implication (1).






share|cite|improve this answer









$endgroup$




... I have started the proof like this:



$forall epsilon>0, exists delta$ so if $delta>0$, $|frac{ln{x}^2}{x}-0|<epsilon$.




The $delta$ in your attempt has nothing to do with $epsilon$.



By definition, given $epsilon>0$, you need to show that there exists $M>0$ such that
$$
x>M quad textbf{implies }quad |frac{2ln x}{x}|<epsilontag{1}
$$



Now for $x>1$,
$$
ln x=int_1^xfrac{1}{t} dtleq int_1^xfrac{1}{sqrt{t}} dt=2sqrt{t}big|_1^x
< 2sqrt{x},
$$
which implies that
$$
frac{2ln x}{x}<frac{4sqrt{x}}{x}=frac{4}{sqrt{x}},quad x>1.
$$
So you want
$
frac{4}{sqrt{x}}<epsilon,
$
which is true if
$$
x>(4/epsilon)^2.
$$
Now, setting
$M=(4/epsilon)^2$, you have the implication (1).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 29 '18 at 4:55







user587192















  • 3




    $begingroup$
    Clever way of showing $ln x<2sqrt x$ for $x>1$.
    $endgroup$
    – DanielWainfleet
    Dec 29 '18 at 6:39










  • $begingroup$
    What does $|_1^x$ mean?
    $endgroup$
    – limitsandlogs224
    Dec 31 '18 at 1:12










  • $begingroup$
    @limitsandlogs224: $f(x)|_a^b:=f(b)-f(a)$.
    $endgroup$
    – user587192
    Dec 31 '18 at 1:53














  • 3




    $begingroup$
    Clever way of showing $ln x<2sqrt x$ for $x>1$.
    $endgroup$
    – DanielWainfleet
    Dec 29 '18 at 6:39










  • $begingroup$
    What does $|_1^x$ mean?
    $endgroup$
    – limitsandlogs224
    Dec 31 '18 at 1:12










  • $begingroup$
    @limitsandlogs224: $f(x)|_a^b:=f(b)-f(a)$.
    $endgroup$
    – user587192
    Dec 31 '18 at 1:53








3




3




$begingroup$
Clever way of showing $ln x<2sqrt x$ for $x>1$.
$endgroup$
– DanielWainfleet
Dec 29 '18 at 6:39




$begingroup$
Clever way of showing $ln x<2sqrt x$ for $x>1$.
$endgroup$
– DanielWainfleet
Dec 29 '18 at 6:39












$begingroup$
What does $|_1^x$ mean?
$endgroup$
– limitsandlogs224
Dec 31 '18 at 1:12




$begingroup$
What does $|_1^x$ mean?
$endgroup$
– limitsandlogs224
Dec 31 '18 at 1:12












$begingroup$
@limitsandlogs224: $f(x)|_a^b:=f(b)-f(a)$.
$endgroup$
– user587192
Dec 31 '18 at 1:53




$begingroup$
@limitsandlogs224: $f(x)|_a^b:=f(b)-f(a)$.
$endgroup$
– user587192
Dec 31 '18 at 1:53











4












$begingroup$

$ln x<2sqrt{x}$ so
$$left|dfrac{ln x^2}{x}right|<dfrac{4}{sqrt{x}}<dfrac{4}{sqrt{M}}leqvarepsilon$$
where $x>M$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How do you prove $ln x< 2sqrt{x}$?
    $endgroup$
    – user587192
    Dec 29 '18 at 2:52










  • $begingroup$
    $ln x<dfrac{x^alpha}{alpha}$ @user587192 with $alpha>0$.
    $endgroup$
    – Nosrati
    Dec 29 '18 at 3:01












  • $begingroup$
    Proof: $e^u>u$ therefore $u>ln u$ let $u=x^alpha$.
    $endgroup$
    – Nosrati
    Dec 29 '18 at 3:15










  • $begingroup$
    @Nosrati, thank you. Could you please explain how I could incorporate $frac{4}{x^{frac{1}{2}}}$ and $frac{4}{M^{frac{1}{2}}}$ into the proof? I thought that I had to solve $frac{4}{x{frac{1}{2}}} <epsilon$ for $x$ so any $epsilon$ is $>|frac{ln{x}^2}{x}-0|$ as long as $delta>frac{16}{epsilon^2}$. I see that they keep $x$ from getting eliminated but I just don't understand how those expressions get substituted into the inequality.
    $endgroup$
    – limitsandlogs224
    Dec 29 '18 at 3:45








  • 1




    $begingroup$
    In fact, it is not difficult to show that $log(x)<sqrt x$ for all $x>0$. Going one step further $log(x)le x^{1/e}$ for all $x$ with equality at $e^e$.
    $endgroup$
    – Mark Viola
    Dec 29 '18 at 20:20


















4












$begingroup$

$ln x<2sqrt{x}$ so
$$left|dfrac{ln x^2}{x}right|<dfrac{4}{sqrt{x}}<dfrac{4}{sqrt{M}}leqvarepsilon$$
where $x>M$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How do you prove $ln x< 2sqrt{x}$?
    $endgroup$
    – user587192
    Dec 29 '18 at 2:52










  • $begingroup$
    $ln x<dfrac{x^alpha}{alpha}$ @user587192 with $alpha>0$.
    $endgroup$
    – Nosrati
    Dec 29 '18 at 3:01












  • $begingroup$
    Proof: $e^u>u$ therefore $u>ln u$ let $u=x^alpha$.
    $endgroup$
    – Nosrati
    Dec 29 '18 at 3:15










  • $begingroup$
    @Nosrati, thank you. Could you please explain how I could incorporate $frac{4}{x^{frac{1}{2}}}$ and $frac{4}{M^{frac{1}{2}}}$ into the proof? I thought that I had to solve $frac{4}{x{frac{1}{2}}} <epsilon$ for $x$ so any $epsilon$ is $>|frac{ln{x}^2}{x}-0|$ as long as $delta>frac{16}{epsilon^2}$. I see that they keep $x$ from getting eliminated but I just don't understand how those expressions get substituted into the inequality.
    $endgroup$
    – limitsandlogs224
    Dec 29 '18 at 3:45








  • 1




    $begingroup$
    In fact, it is not difficult to show that $log(x)<sqrt x$ for all $x>0$. Going one step further $log(x)le x^{1/e}$ for all $x$ with equality at $e^e$.
    $endgroup$
    – Mark Viola
    Dec 29 '18 at 20:20
















4












4








4





$begingroup$

$ln x<2sqrt{x}$ so
$$left|dfrac{ln x^2}{x}right|<dfrac{4}{sqrt{x}}<dfrac{4}{sqrt{M}}leqvarepsilon$$
where $x>M$.






share|cite|improve this answer









$endgroup$



$ln x<2sqrt{x}$ so
$$left|dfrac{ln x^2}{x}right|<dfrac{4}{sqrt{x}}<dfrac{4}{sqrt{M}}leqvarepsilon$$
where $x>M$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 29 '18 at 2:40









NosratiNosrati

26.5k62354




26.5k62354












  • $begingroup$
    How do you prove $ln x< 2sqrt{x}$?
    $endgroup$
    – user587192
    Dec 29 '18 at 2:52










  • $begingroup$
    $ln x<dfrac{x^alpha}{alpha}$ @user587192 with $alpha>0$.
    $endgroup$
    – Nosrati
    Dec 29 '18 at 3:01












  • $begingroup$
    Proof: $e^u>u$ therefore $u>ln u$ let $u=x^alpha$.
    $endgroup$
    – Nosrati
    Dec 29 '18 at 3:15










  • $begingroup$
    @Nosrati, thank you. Could you please explain how I could incorporate $frac{4}{x^{frac{1}{2}}}$ and $frac{4}{M^{frac{1}{2}}}$ into the proof? I thought that I had to solve $frac{4}{x{frac{1}{2}}} <epsilon$ for $x$ so any $epsilon$ is $>|frac{ln{x}^2}{x}-0|$ as long as $delta>frac{16}{epsilon^2}$. I see that they keep $x$ from getting eliminated but I just don't understand how those expressions get substituted into the inequality.
    $endgroup$
    – limitsandlogs224
    Dec 29 '18 at 3:45








  • 1




    $begingroup$
    In fact, it is not difficult to show that $log(x)<sqrt x$ for all $x>0$. Going one step further $log(x)le x^{1/e}$ for all $x$ with equality at $e^e$.
    $endgroup$
    – Mark Viola
    Dec 29 '18 at 20:20




















  • $begingroup$
    How do you prove $ln x< 2sqrt{x}$?
    $endgroup$
    – user587192
    Dec 29 '18 at 2:52










  • $begingroup$
    $ln x<dfrac{x^alpha}{alpha}$ @user587192 with $alpha>0$.
    $endgroup$
    – Nosrati
    Dec 29 '18 at 3:01












  • $begingroup$
    Proof: $e^u>u$ therefore $u>ln u$ let $u=x^alpha$.
    $endgroup$
    – Nosrati
    Dec 29 '18 at 3:15










  • $begingroup$
    @Nosrati, thank you. Could you please explain how I could incorporate $frac{4}{x^{frac{1}{2}}}$ and $frac{4}{M^{frac{1}{2}}}$ into the proof? I thought that I had to solve $frac{4}{x{frac{1}{2}}} <epsilon$ for $x$ so any $epsilon$ is $>|frac{ln{x}^2}{x}-0|$ as long as $delta>frac{16}{epsilon^2}$. I see that they keep $x$ from getting eliminated but I just don't understand how those expressions get substituted into the inequality.
    $endgroup$
    – limitsandlogs224
    Dec 29 '18 at 3:45








  • 1




    $begingroup$
    In fact, it is not difficult to show that $log(x)<sqrt x$ for all $x>0$. Going one step further $log(x)le x^{1/e}$ for all $x$ with equality at $e^e$.
    $endgroup$
    – Mark Viola
    Dec 29 '18 at 20:20


















$begingroup$
How do you prove $ln x< 2sqrt{x}$?
$endgroup$
– user587192
Dec 29 '18 at 2:52




$begingroup$
How do you prove $ln x< 2sqrt{x}$?
$endgroup$
– user587192
Dec 29 '18 at 2:52












$begingroup$
$ln x<dfrac{x^alpha}{alpha}$ @user587192 with $alpha>0$.
$endgroup$
– Nosrati
Dec 29 '18 at 3:01






$begingroup$
$ln x<dfrac{x^alpha}{alpha}$ @user587192 with $alpha>0$.
$endgroup$
– Nosrati
Dec 29 '18 at 3:01














$begingroup$
Proof: $e^u>u$ therefore $u>ln u$ let $u=x^alpha$.
$endgroup$
– Nosrati
Dec 29 '18 at 3:15




$begingroup$
Proof: $e^u>u$ therefore $u>ln u$ let $u=x^alpha$.
$endgroup$
– Nosrati
Dec 29 '18 at 3:15












$begingroup$
@Nosrati, thank you. Could you please explain how I could incorporate $frac{4}{x^{frac{1}{2}}}$ and $frac{4}{M^{frac{1}{2}}}$ into the proof? I thought that I had to solve $frac{4}{x{frac{1}{2}}} <epsilon$ for $x$ so any $epsilon$ is $>|frac{ln{x}^2}{x}-0|$ as long as $delta>frac{16}{epsilon^2}$. I see that they keep $x$ from getting eliminated but I just don't understand how those expressions get substituted into the inequality.
$endgroup$
– limitsandlogs224
Dec 29 '18 at 3:45






$begingroup$
@Nosrati, thank you. Could you please explain how I could incorporate $frac{4}{x^{frac{1}{2}}}$ and $frac{4}{M^{frac{1}{2}}}$ into the proof? I thought that I had to solve $frac{4}{x{frac{1}{2}}} <epsilon$ for $x$ so any $epsilon$ is $>|frac{ln{x}^2}{x}-0|$ as long as $delta>frac{16}{epsilon^2}$. I see that they keep $x$ from getting eliminated but I just don't understand how those expressions get substituted into the inequality.
$endgroup$
– limitsandlogs224
Dec 29 '18 at 3:45






1




1




$begingroup$
In fact, it is not difficult to show that $log(x)<sqrt x$ for all $x>0$. Going one step further $log(x)le x^{1/e}$ for all $x$ with equality at $e^e$.
$endgroup$
– Mark Viola
Dec 29 '18 at 20:20






$begingroup$
In fact, it is not difficult to show that $log(x)<sqrt x$ for all $x>0$. Going one step further $log(x)le x^{1/e}$ for all $x$ with equality at $e^e$.
$endgroup$
– Mark Viola
Dec 29 '18 at 20:20




















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