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I have this limit :

$\displaystyle \lim_{x\to 0} \left(\dfrac{1}{\ln(x+\sqrt{x^2+1})} - \dfrac{1}{\ln(x+1)}\right) = -\dfrac{1}{2}$

it seems that this value is found by using L'Hopital rule, but the result from using it seems very messy and chaotic.

My question is :

  • Is there any way i can evaluate this limit without L'Hopital ?

So far :

  • I have tried Hyperbolic Subtitution (by letting $x = \sinh(t), t\to 0$,

    and thus imply that $x+\sqrt{x^2+1} = \sinh(t)+\cosh(t) = e^t$, and since $\sinh(t) 0$ for any t value approaching $0$, means that $\sinh(t) + 1 \approx \cosh(t)$, there's still no luck because i still have to use L'Hopital (and even with it i still get indeterminate form). I also tried not to change to $\sinh(t)+1$ into $\cosh(t)$, still also need L'hopitals too.

Any help is appreciated.

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    $\begingroup$ Personally I would just use the series expansions of these functions: the denominator of the first fraction is $x+O(x^3)$, of the second is $x-\frac{1}{2}x^2+O(x^3)$, so the whole thing is $-\frac12+O(x)$. $\endgroup$ Jun 25, 2021 at 8:07

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$$\dfrac{1}{\ln(x+\sqrt{x^2+1})} - \dfrac{1}{\ln(x+1)}$$

Using series and then long division $$\ln(x+\sqrt{x^2+1})=x-\frac{x^3}{6}+O\left(x^5\right)\implies \dfrac{1}{\ln(x+\sqrt{x^2+1})}=\frac{1}{x}+\frac{x}{6}+O\left(x^3\right)$$ $$\ln(x+1)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+O\left(x^5\right)\implies \dfrac{1}{\ln(x+1)}=\frac{1}{x}+\frac{1}{2}-\frac{x}{12}+\frac{x^2}{24}+O\left(x^3\right)$$

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  • $\begingroup$ Thanks, i was so absorbed in working the limit that i forget to consider T-Series. $\endgroup$
    – JangoHypno
    Jun 25, 2021 at 8:26
  • $\begingroup$ Developing directly $\ln(x+\sqrt(1+x^2))$ is not difficult, but it might help to notice that $\ln(x+\sqrt(1+x^2))=\mathrm{argsh}\,x$, as it's often one of the "known" series. And even if not, $\mathrm{argsh}'\,x=\frac{1}{\sqrt{1+x^2}}$ could also help to find the derivation. There is more than one way to achieve this :) $\endgroup$ Jun 25, 2021 at 8:37
  • $\begingroup$ @Jean-ClaudeArbaut. There are many ways as you wrote. It is a long time I did not see $\mathrm{argsh}(x)$ which is what I learnt ... 64 years agot ! Cheers :-) $\endgroup$ Jun 25, 2021 at 8:41
  • $\begingroup$ Brilliant !! another bad thing being too absorbed about hahaha (though i think you should add a little bit detail that $\text{sh}$ mean $\sinh$, or just use $\sinh^{-1}$ ). $\endgroup$
    – JangoHypno
    Jun 25, 2021 at 8:43
  • $\begingroup$ @JamboRambo Yes, it's the notation used in France. $\mathrm{sh}\,x=\mathrm{sinh}\,x$ and $\mathrm{argsh}\,x=\mathrm{arsinh}\,x=\mathrm{sinh^{-1}}\,x$. sinh and arsinh are recommended by the ISO 80000-2 norm. $\endgroup$ Jun 25, 2021 at 8:45
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You can use the hyperbolic substitution partly. This makes L'Hospital much more manageable:

$$\lim_{x\to 0} \left(\dfrac{1}{\ln(x+\sqrt{x^2+1})} - \dfrac{1}{\ln(x+1)}\right) \\=\lim_{x\to 0} \left(\dfrac{1}{\ln(x+\sqrt{x^2+1})} - \dfrac{1}x\right)+\lim_{x\to 0} \left(\dfrac{1}x - \dfrac{1}{\ln(x+1)}\right) \\=\lim_{t\to 0} \left(\dfrac{1}t - \dfrac{1}{\sinh(t)}\right)+\lim_{x\to 0} \left(\dfrac{1}x - \dfrac{1}{\ln(x+1)}\right).$$


$$\dfrac{1}t - \dfrac{1}{\sinh(t)}=\frac{\sinh(t)-t}{t^2}\frac{t}{\sinh(t)}\to\frac{\cosh(t)-1}{2t}\cdot1\to\frac{\sinh(t)}2\to0.$$

or $$\dfrac{t-\dfrac{t^3}{3!}+\dfrac{t^5}{5!}+\cdots-t}{t\sinh(t)}\to0$$


$$\frac1x-\frac1{\log(x+1)}=\frac{\log(x+1)-x}{x^2}\frac{x}{\log(x+1)}\to\frac{-\dfrac x{x+1}}{2x}\cdot1\to-\frac12.$$ or

$$\frac{x-\dfrac{x^2}2+\dfrac{x^3}3-\cdots-x}{x\log(x+1)}\to-\frac12$$


We used another trick, to simplify the denominators and avoid painful differentiations.

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    $\begingroup$ On $\dfrac{\sinh(t) - t}{t^2} \dfrac{t}{\sinh(t)}$ did you just use L'Hopital on two different limit ($\dfrac{\sinh(t) - t}{t^2}$ and another one on $\dfrac{t}{\sinh(t)}$ ) ? $\endgroup$
    – JangoHypno
    Jun 25, 2021 at 9:10
  • $\begingroup$ @JamboRambo: I used $\dfrac t{\sinh(t)}\to1$ directly (though L'Hospital works as well). I also used $\dfrac x{\log(x+1)}\to1$. $\endgroup$
    – user65203
    Jun 25, 2021 at 9:13
  • $\begingroup$ Oh ok, so it works on Hyperbolic one too. $\endgroup$
    – JangoHypno
    Jun 25, 2021 at 9:15
  • $\begingroup$ @JamboRambo: yep, $x\sim\sin x\sim\tan x\sim\sinh x\sim\tanh x\sim\log(1+x)\sim e^x-1\cdots$. $\endgroup$
    – user65203
    Jun 25, 2021 at 9:16
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Let's define $$L=\displaystyle \lim_{x\to 0} \left(\dfrac{1}{\ln(x+\sqrt{x^2+1})} - \dfrac{1}{\ln(x+1)}\right) = \lim_{x \to 0} \left(\dfrac{1}{\ln \left(1+\left(x+\sqrt{x^2+1}-1 \right) \right) } - \dfrac{1}{\ln(x+1)}\right)$$ Then as $x \to 0$ $$\frac{1}{\ln \left(1 +\left(\sqrt{x^2+1} +x-1 \right) \right)} \sim \left(\frac{1}{x+\sqrt{x^2+1}-1} \right) \sim \frac{1}{x+x^2/2} $$ Then we've $$L=\lim_{x \to 0}\left(\frac{1}{x+x^2/2}-\frac{1}{\ln(1+x)} \right) =\lim_{x \to 0}\left(\frac{\ln(1+x)-x-x^2/2}{x(1+x/2)\ln(1+x)} \right)$$ $$\sim \frac{-x^2 /2}{x^2 (1+x/2)} \to \frac{-1}{2}$$ as $x \to 0$

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  • $\begingroup$ The $\sim$ has a precise meaning, and it's not true that if $A\sim a$ and $B\sim b$ then automatically $A-B\sim a-b$. Here you have proved nothing. $\endgroup$ Jun 25, 2021 at 8:43
  • $\begingroup$ Thanks for the comment, I'm aware that if we write $f(x) \sim g(x)$ as $x \to a$ , then we must have $\lim_{x \to a}\frac{f(x)}{g(x)} =1$ , I see where I got it wrong, I've made some corrections. Please comment in case I'm still missing the necessary rigour. Thanks $\endgroup$
    – A S D
    Jun 25, 2021 at 9:40
  • $\begingroup$ It doesn't work either. You could as well have written $\frac{1}{x+\sqrt{x^2+1}-1} \sim \frac{1}{x}+a$ for any constant $a$ (therefore find out that the limit is $a-1/2$, which is of course wrong). And when you replace $\frac{1}{x+\sqrt{x^2+1}-1}$ with $\frac{1}{x+x^2/2}$ in the limit below, it's wrong for the same reason as in my first comment. $\endgroup$ Jun 25, 2021 at 10:08
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Let $x = \sinh(t)$, since $\cosh^{2}(t)-\sinh^2(t) = 1 $ then $x^2+1 = \cosh^2(x)$ and $x+\sqrt{x^2+1} = \sinh(x)+\cosh(x)$.

Since $\sinh(t) = \dfrac{1}{2} \cdot \Delta\exp(t,-t)$ and $\cosh(x) = \overline{\exp} (t,-t)$ (the bar mean average), then $\sinh(t)+\cos(t) = e^t$

$L = \displaystyle \lim_{x\to 0} \left(\dfrac{1}{\ln(x+\sqrt{x^2+1})} - \dfrac{1}{\ln(x+1)}\right) = \lim_{t\to 0} \left(\dfrac{1}{\ln(e^t)} - \dfrac{1}{\ln(\sinh(x)+1)}\right)$

Since $\lim_{t\to 0} \sinh(t) = 0$ means that $\lim_{t\to 0} \sinh(t)+1 = 1 = \lim_{t\to 0} \cosh(t)$ and i can subtitute $\sinh(t) + 1 = \cosh(t)$

$L = \displaystyle \lim_{x\to 0} \left(\dfrac{1}{\ln(x+\sqrt{x^2+1})} - \dfrac{1}{\ln(x+1)}\right) = \lim_{t\to 0} \left(\dfrac{1}{t} - \dfrac{1}{\ln(\cosh(t))}\right)$

$\displaystyle L = \lim_{t\to 0} \left(\dfrac{\ln(\cosh(t))-t}{t\ln(\cosh(t))}\right)$

and i'm stuck on this, need to use L'Hopital afterward.

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  • $\begingroup$ Note that replacing $\sinh t +1$ by $\cosh t$ may lead to a false result as $\sinh t+1 =1+t+\mathcal{O}(t^3)$ and $\cosh t = 1+t^2/2+\mathcal{O}(t^4)$. So the latter goes faster to $1$ than the former. $\endgroup$
    – Gary
    Jun 25, 2021 at 8:18
  • $\begingroup$ You could have put this in your post, as an edit. $\endgroup$ Jun 25, 2021 at 8:19
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Just using fundamental limits $$\frac{\log(1+\alpha(x))}{\alpha(x)} \to 1$$ and $$\frac{(1+\alpha(x))^k-1}{\alpha(x)}\to k$$ for $\alpha(x) \to 0$, one obtains \begin{eqnarray} \mathcal L &=& \lim_{x\to 0}\left(\frac1{\log(x+\sqrt{x^2+1})}-\frac1{\log(x+1)}\right)=\\ &=&\lim_{x\to 0}\frac{\log\frac{x+1}{x+\sqrt{x^2+1}}}{\log(x+\sqrt{x^2+1})\log(x+1)}=\\ &=&\lim_{x\to 0}\frac{\frac{x+1}{x+\sqrt{x^2+1}}-1}{x(x+\sqrt{x^2+ 1}-1)}=\\ &=&\lim_{x\to 0}\frac{x+1-x-\sqrt{x^2+1}}{x(x+\sqrt{x^2+1})(x+\sqrt{x^2+ 1}-1)}=\\ &=&\lim_{x\to 0}\frac{1-\sqrt{x^2+1}}{x(x-1+\sqrt{x^2+ 1})}=\\ &=&-\frac12 \lim_{x\to 0}\frac{x}{x-1+\sqrt{x^2+1}}=\\ &=&-\frac12 \lim_{x\to 0}\frac{x(x-1-\sqrt{x^2+1})}{x^2-2x+1-x^2-1}=\\ &=&-\frac12 \lim_{x\to 0}\frac{-2x}{-2x}=-\frac12. \end{eqnarray}

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