Calculate $lim_{x to 0} frac{ln(1+2x)}{x^2}$ with the help of l'Hospital's and Bernoullie's rule.












3












$begingroup$



Task:



Calculate $$lim_{x to 0} frac{ln(1+2x)}{x^2}$$ with the help of
l'Hospital's and Bernoullie's rule.




My thoughts:



Because $mathcal{D}(f)={xmid xinmathbb{R} land xneq 0}$ the function is undefined for $0$ and therefore, I need to find out, whether the function has a limit or only one-sided limits. In order to do that, I'll just calculate the one sided limits. If $$lim_{x to 0^+} frac{ln(1+2x)}{x^2}neq lim_{x to 0^-} frac{ln(1+2x)}{x^2} implies lim_{x to 0} frac{ln(1+2x)}{x^2} text{ doesn't exist}$$



$lim_{x to 0^-} frac{ln(1+2x)}{x^2}overbrace{=}^{l'Hospital}=lim_{x to 0^-} frac{2/(2x+1)}{2x}=lim_{x to 0^-}frac{1}{x(2x+1)}overbrace{=}^{product- rule}underbrace{lim_{x to 0^-}frac{1}{(2x+1)}}_{=1}cdot underbrace{lim_{x to 0^-}frac{1}{x}}_{(*)}=1cdot (*)=(*)$



$(*)$: If $x$ is small, than $1/x$ gets proportional bigger. Let $M>0$ and let $delta = 1/M$. Than $-1/x<frac{-1}{1/M}=-M;forall (-delta)<x<0$. Since $M$ can be arbitrarily large: $$lim_{x to 0^-} frac1x=-infty$$



$lim_{x to 0^-}$ analogue. $$lim_{x to 0^+} frac{ln(1+2x)}{x^2} = cdots = lim_{x to 0^+} frac1x=infty$$ $implies lim_{x to 0}$ doesn't exist.



Is this proof correct?






real-analysis calculus proof-verification







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes the searched limit doesn't exist.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 10 '18 at 20:16










  • $begingroup$
    Can you use Maclaurin series? Logarithm expansion immediately gives you the resut
    $endgroup$
    – Alex
    Dec 10 '18 at 20:23






  • 1




    $begingroup$
    @Alex I've never heard of the Maclaurin Series, but I'm curious how that proof would look like! Please share it.
    $endgroup$
    – Doesbaddel
    Dec 10 '18 at 20:25








  • 2




    $begingroup$
    Maybe you have heard of Taylor series? It’s practically the same thing. Actually I don’t know why they are called differently, anybody knows?
    $endgroup$
    – tommy1996q
    Dec 10 '18 at 20:30








  • 2




    $begingroup$
    @tommy1996q It's just nomenclature. A Maclaurin series is just a Taylor expansion about $x=0$.
    $endgroup$
    – bob.sacamento
    Dec 10 '18 at 20:35
















3












$begingroup$



Task:



Calculate $$lim_{x to 0} frac{ln(1+2x)}{x^2}$$ with the help of
l'Hospital's and Bernoullie's rule.




My thoughts:



Because $mathcal{D}(f)={xmid xinmathbb{R} land xneq 0}$ the function is undefined for $0$ and therefore, I need to find out, whether the function has a limit or only one-sided limits. In order to do that, I'll just calculate the one sided limits. If $$lim_{x to 0^+} frac{ln(1+2x)}{x^2}neq lim_{x to 0^-} frac{ln(1+2x)}{x^2} implies lim_{x to 0} frac{ln(1+2x)}{x^2} text{ doesn't exist}$$



$lim_{x to 0^-} frac{ln(1+2x)}{x^2}overbrace{=}^{l'Hospital}=lim_{x to 0^-} frac{2/(2x+1)}{2x}=lim_{x to 0^-}frac{1}{x(2x+1)}overbrace{=}^{product- rule}underbrace{lim_{x to 0^-}frac{1}{(2x+1)}}_{=1}cdot underbrace{lim_{x to 0^-}frac{1}{x}}_{(*)}=1cdot (*)=(*)$



$(*)$: If $x$ is small, than $1/x$ gets proportional bigger. Let $M>0$ and let $delta = 1/M$. Than $-1/x<frac{-1}{1/M}=-M;forall (-delta)<x<0$. Since $M$ can be arbitrarily large: $$lim_{x to 0^-} frac1x=-infty$$



$lim_{x to 0^-}$ analogue. $$lim_{x to 0^+} frac{ln(1+2x)}{x^2} = cdots = lim_{x to 0^+} frac1x=infty$$ $implies lim_{x to 0}$ doesn't exist.



Is this proof correct?






real-analysis calculus proof-verification







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes the searched limit doesn't exist.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 10 '18 at 20:16










  • $begingroup$
    Can you use Maclaurin series? Logarithm expansion immediately gives you the resut
    $endgroup$
    – Alex
    Dec 10 '18 at 20:23






  • 1




    $begingroup$
    @Alex I've never heard of the Maclaurin Series, but I'm curious how that proof would look like! Please share it.
    $endgroup$
    – Doesbaddel
    Dec 10 '18 at 20:25








  • 2




    $begingroup$
    Maybe you have heard of Taylor series? It’s practically the same thing. Actually I don’t know why they are called differently, anybody knows?
    $endgroup$
    – tommy1996q
    Dec 10 '18 at 20:30








  • 2




    $begingroup$
    @tommy1996q It's just nomenclature. A Maclaurin series is just a Taylor expansion about $x=0$.
    $endgroup$
    – bob.sacamento
    Dec 10 '18 at 20:35














3












3








3





$begingroup$



Task:



Calculate $$lim_{x to 0} frac{ln(1+2x)}{x^2}$$ with the help of
l'Hospital's and Bernoullie's rule.




My thoughts:



Because $mathcal{D}(f)={xmid xinmathbb{R} land xneq 0}$ the function is undefined for $0$ and therefore, I need to find out, whether the function has a limit or only one-sided limits. In order to do that, I'll just calculate the one sided limits. If $$lim_{x to 0^+} frac{ln(1+2x)}{x^2}neq lim_{x to 0^-} frac{ln(1+2x)}{x^2} implies lim_{x to 0} frac{ln(1+2x)}{x^2} text{ doesn't exist}$$



$lim_{x to 0^-} frac{ln(1+2x)}{x^2}overbrace{=}^{l'Hospital}=lim_{x to 0^-} frac{2/(2x+1)}{2x}=lim_{x to 0^-}frac{1}{x(2x+1)}overbrace{=}^{product- rule}underbrace{lim_{x to 0^-}frac{1}{(2x+1)}}_{=1}cdot underbrace{lim_{x to 0^-}frac{1}{x}}_{(*)}=1cdot (*)=(*)$



$(*)$: If $x$ is small, than $1/x$ gets proportional bigger. Let $M>0$ and let $delta = 1/M$. Than $-1/x<frac{-1}{1/M}=-M;forall (-delta)<x<0$. Since $M$ can be arbitrarily large: $$lim_{x to 0^-} frac1x=-infty$$



$lim_{x to 0^-}$ analogue. $$lim_{x to 0^+} frac{ln(1+2x)}{x^2} = cdots = lim_{x to 0^+} frac1x=infty$$ $implies lim_{x to 0}$ doesn't exist.



Is this proof correct?






real-analysis calculus proof-verification







share|cite|improve this question











$endgroup$





Task:



Calculate $$lim_{x to 0} frac{ln(1+2x)}{x^2}$$ with the help of
l'Hospital's and Bernoullie's rule.




My thoughts:



Because $mathcal{D}(f)={xmid xinmathbb{R} land xneq 0}$ the function is undefined for $0$ and therefore, I need to find out, whether the function has a limit or only one-sided limits. In order to do that, I'll just calculate the one sided limits. If $$lim_{x to 0^+} frac{ln(1+2x)}{x^2}neq lim_{x to 0^-} frac{ln(1+2x)}{x^2} implies lim_{x to 0} frac{ln(1+2x)}{x^2} text{ doesn't exist}$$



$lim_{x to 0^-} frac{ln(1+2x)}{x^2}overbrace{=}^{l'Hospital}=lim_{x to 0^-} frac{2/(2x+1)}{2x}=lim_{x to 0^-}frac{1}{x(2x+1)}overbrace{=}^{product- rule}underbrace{lim_{x to 0^-}frac{1}{(2x+1)}}_{=1}cdot underbrace{lim_{x to 0^-}frac{1}{x}}_{(*)}=1cdot (*)=(*)$



$(*)$: If $x$ is small, than $1/x$ gets proportional bigger. Let $M>0$ and let $delta = 1/M$. Than $-1/x<frac{-1}{1/M}=-M;forall (-delta)<x<0$. Since $M$ can be arbitrarily large: $$lim_{x to 0^-} frac1x=-infty$$



$lim_{x to 0^-}$ analogue. $$lim_{x to 0^+} frac{ln(1+2x)}{x^2} = cdots = lim_{x to 0^+} frac1x=infty$$ $implies lim_{x to 0}$ doesn't exist.



Is this proof correct?






real-analysis calculus proof-verification




real-analysis calculus proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 20:27







Doesbaddel

















asked Dec 10 '18 at 20:13









DoesbaddelDoesbaddel

34011




34011








  • 1




    $begingroup$
    Yes the searched limit doesn't exist.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 10 '18 at 20:16










  • $begingroup$
    Can you use Maclaurin series? Logarithm expansion immediately gives you the resut
    $endgroup$
    – Alex
    Dec 10 '18 at 20:23






  • 1




    $begingroup$
    @Alex I've never heard of the Maclaurin Series, but I'm curious how that proof would look like! Please share it.
    $endgroup$
    – Doesbaddel
    Dec 10 '18 at 20:25








  • 2




    $begingroup$
    Maybe you have heard of Taylor series? It’s practically the same thing. Actually I don’t know why they are called differently, anybody knows?
    $endgroup$
    – tommy1996q
    Dec 10 '18 at 20:30








  • 2




    $begingroup$
    @tommy1996q It's just nomenclature. A Maclaurin series is just a Taylor expansion about $x=0$.
    $endgroup$
    – bob.sacamento
    Dec 10 '18 at 20:35














  • 1




    $begingroup$
    Yes the searched limit doesn't exist.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 10 '18 at 20:16










  • $begingroup$
    Can you use Maclaurin series? Logarithm expansion immediately gives you the resut
    $endgroup$
    – Alex
    Dec 10 '18 at 20:23






  • 1




    $begingroup$
    @Alex I've never heard of the Maclaurin Series, but I'm curious how that proof would look like! Please share it.
    $endgroup$
    – Doesbaddel
    Dec 10 '18 at 20:25








  • 2




    $begingroup$
    Maybe you have heard of Taylor series? It’s practically the same thing. Actually I don’t know why they are called differently, anybody knows?
    $endgroup$
    – tommy1996q
    Dec 10 '18 at 20:30








  • 2




    $begingroup$
    @tommy1996q It's just nomenclature. A Maclaurin series is just a Taylor expansion about $x=0$.
    $endgroup$
    – bob.sacamento
    Dec 10 '18 at 20:35








1




1




$begingroup$
Yes the searched limit doesn't exist.
$endgroup$
– Dr. Sonnhard Graubner
Dec 10 '18 at 20:16




$begingroup$
Yes the searched limit doesn't exist.
$endgroup$
– Dr. Sonnhard Graubner
Dec 10 '18 at 20:16












$begingroup$
Can you use Maclaurin series? Logarithm expansion immediately gives you the resut
$endgroup$
– Alex
Dec 10 '18 at 20:23




$begingroup$
Can you use Maclaurin series? Logarithm expansion immediately gives you the resut
$endgroup$
– Alex
Dec 10 '18 at 20:23




1




1




$begingroup$
@Alex I've never heard of the Maclaurin Series, but I'm curious how that proof would look like! Please share it.
$endgroup$
– Doesbaddel
Dec 10 '18 at 20:25






$begingroup$
@Alex I've never heard of the Maclaurin Series, but I'm curious how that proof would look like! Please share it.
$endgroup$
– Doesbaddel
Dec 10 '18 at 20:25






2




2




$begingroup$
Maybe you have heard of Taylor series? It’s practically the same thing. Actually I don’t know why they are called differently, anybody knows?
$endgroup$
– tommy1996q
Dec 10 '18 at 20:30






$begingroup$
Maybe you have heard of Taylor series? It’s practically the same thing. Actually I don’t know why they are called differently, anybody knows?
$endgroup$
– tommy1996q
Dec 10 '18 at 20:30






2




2




$begingroup$
@tommy1996q It's just nomenclature. A Maclaurin series is just a Taylor expansion about $x=0$.
$endgroup$
– bob.sacamento
Dec 10 '18 at 20:35




$begingroup$
@tommy1996q It's just nomenclature. A Maclaurin series is just a Taylor expansion about $x=0$.
$endgroup$
– bob.sacamento
Dec 10 '18 at 20:35










2 Answers
2






active

oldest

votes


















2












$begingroup$

Yes your evaluation is fine, to check it by standard limits, we have that



$$ frac{ln(1+2x)}{x^2}=frac{ln(1+2x)}{2x}frac{2}{x}to 1cdotpminfty$$



therefore the limit doesn't exist.



For the proof of the standard limit refer to




  • Determine the following limit as x approaches 0: $frac{ln(1+x)}x$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much, that shortens the proof significantly!
    $endgroup$
    – Doesbaddel
    Dec 11 '18 at 18:14






  • 1




    $begingroup$
    @Doesbaddel You are welcome! That's strictly related to some general suggestions given here
    $endgroup$
    – gimusi
    Dec 11 '18 at 18:15



















2












$begingroup$

Since as $x to 0 log (1 + 2x) to 0$, you can expand $log$ function around $x=0$ to get (first term is enough) $log (1+2x) sim 2x$, and the fraction becomes $frac{2}{x}$ that certainly diverges






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, way shorter than my thoughts.
    $endgroup$
    – Doesbaddel
    Dec 11 '18 at 18:13











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Yes your evaluation is fine, to check it by standard limits, we have that



$$ frac{ln(1+2x)}{x^2}=frac{ln(1+2x)}{2x}frac{2}{x}to 1cdotpminfty$$



therefore the limit doesn't exist.



For the proof of the standard limit refer to




  • Determine the following limit as x approaches 0: $frac{ln(1+x)}x$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much, that shortens the proof significantly!
    $endgroup$
    – Doesbaddel
    Dec 11 '18 at 18:14






  • 1




    $begingroup$
    @Doesbaddel You are welcome! That's strictly related to some general suggestions given here
    $endgroup$
    – gimusi
    Dec 11 '18 at 18:15
















2












$begingroup$

Yes your evaluation is fine, to check it by standard limits, we have that



$$ frac{ln(1+2x)}{x^2}=frac{ln(1+2x)}{2x}frac{2}{x}to 1cdotpminfty$$



therefore the limit doesn't exist.



For the proof of the standard limit refer to




  • Determine the following limit as x approaches 0: $frac{ln(1+x)}x$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much, that shortens the proof significantly!
    $endgroup$
    – Doesbaddel
    Dec 11 '18 at 18:14






  • 1




    $begingroup$
    @Doesbaddel You are welcome! That's strictly related to some general suggestions given here
    $endgroup$
    – gimusi
    Dec 11 '18 at 18:15














2












2








2





$begingroup$

Yes your evaluation is fine, to check it by standard limits, we have that



$$ frac{ln(1+2x)}{x^2}=frac{ln(1+2x)}{2x}frac{2}{x}to 1cdotpminfty$$



therefore the limit doesn't exist.



For the proof of the standard limit refer to




  • Determine the following limit as x approaches 0: $frac{ln(1+x)}x$






share|cite|improve this answer









$endgroup$



Yes your evaluation is fine, to check it by standard limits, we have that



$$ frac{ln(1+2x)}{x^2}=frac{ln(1+2x)}{2x}frac{2}{x}to 1cdotpminfty$$



therefore the limit doesn't exist.



For the proof of the standard limit refer to




  • Determine the following limit as x approaches 0: $frac{ln(1+x)}x$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 21:10









gimusigimusi

92.8k84494




92.8k84494












  • $begingroup$
    Thank you very much, that shortens the proof significantly!
    $endgroup$
    – Doesbaddel
    Dec 11 '18 at 18:14






  • 1




    $begingroup$
    @Doesbaddel You are welcome! That's strictly related to some general suggestions given here
    $endgroup$
    – gimusi
    Dec 11 '18 at 18:15


















  • $begingroup$
    Thank you very much, that shortens the proof significantly!
    $endgroup$
    – Doesbaddel
    Dec 11 '18 at 18:14






  • 1




    $begingroup$
    @Doesbaddel You are welcome! That's strictly related to some general suggestions given here
    $endgroup$
    – gimusi
    Dec 11 '18 at 18:15
















$begingroup$
Thank you very much, that shortens the proof significantly!
$endgroup$
– Doesbaddel
Dec 11 '18 at 18:14




$begingroup$
Thank you very much, that shortens the proof significantly!
$endgroup$
– Doesbaddel
Dec 11 '18 at 18:14




1




1




$begingroup$
@Doesbaddel You are welcome! That's strictly related to some general suggestions given here
$endgroup$
– gimusi
Dec 11 '18 at 18:15




$begingroup$
@Doesbaddel You are welcome! That's strictly related to some general suggestions given here
$endgroup$
– gimusi
Dec 11 '18 at 18:15











2












$begingroup$

Since as $x to 0 log (1 + 2x) to 0$, you can expand $log$ function around $x=0$ to get (first term is enough) $log (1+2x) sim 2x$, and the fraction becomes $frac{2}{x}$ that certainly diverges






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, way shorter than my thoughts.
    $endgroup$
    – Doesbaddel
    Dec 11 '18 at 18:13
















2












$begingroup$

Since as $x to 0 log (1 + 2x) to 0$, you can expand $log$ function around $x=0$ to get (first term is enough) $log (1+2x) sim 2x$, and the fraction becomes $frac{2}{x}$ that certainly diverges






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, way shorter than my thoughts.
    $endgroup$
    – Doesbaddel
    Dec 11 '18 at 18:13














2












2








2





$begingroup$

Since as $x to 0 log (1 + 2x) to 0$, you can expand $log$ function around $x=0$ to get (first term is enough) $log (1+2x) sim 2x$, and the fraction becomes $frac{2}{x}$ that certainly diverges






share|cite|improve this answer









$endgroup$



Since as $x to 0 log (1 + 2x) to 0$, you can expand $log$ function around $x=0$ to get (first term is enough) $log (1+2x) sim 2x$, and the fraction becomes $frac{2}{x}$ that certainly diverges







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 21:11









AlexAlex

14.3k42134




14.3k42134












  • $begingroup$
    Thanks, way shorter than my thoughts.
    $endgroup$
    – Doesbaddel
    Dec 11 '18 at 18:13


















  • $begingroup$
    Thanks, way shorter than my thoughts.
    $endgroup$
    – Doesbaddel
    Dec 11 '18 at 18:13
















$begingroup$
Thanks, way shorter than my thoughts.
$endgroup$
– Doesbaddel
Dec 11 '18 at 18:13




$begingroup$
Thanks, way shorter than my thoughts.
$endgroup$
– Doesbaddel
Dec 11 '18 at 18:13


















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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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