Is it Possible to Evaluate This Limit Without L'Hopital Rule?

Question

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.

Answer 1

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

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

Answer 2

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.

Answer 3

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}

Answer 4

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 $x \to 0$

Answer 5

You can use the hyperbolic substitution partly. This makes L'Hospital much more manageable:

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

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

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

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

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

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