Limit of $\left(y+\frac23\right)\mathrm{ln}\left(\frac{\sqrt{1+y}+1}{\sqrt{1+y}-1} \right) - 2 \sqrt{1+y}$ We consider the limit for large $y$ of the following expression :
$$\left(y+\frac23\right)\mathrm{ln}\left(\dfrac{\sqrt{1+y}+1}{\sqrt{1+y}-1} \right) - 2 \sqrt{1+y}.$$
Many references state that the large $y$ behaviour of this last expression is $y^{-3/2}$. I have been trying to show this in any way possible but I'm only hitting dead ends. Wolframalpha gives a Puiseux series of that expression which makes this behaviour explicit, but I do not understand how to get to this series either. Could you please help me on this?
 A: One can write $1+y=1/t^2$ so that $t\to 0^+$ and then the given expression, say $F(y) $, transforms into $$F(y) =\left(\frac{1}{t^2}-\frac{1}{3}\right)\log\frac{1+t}{1-t}-\frac{2}{t}\tag{1}$$ and we need to find the limit of $y^{3/2}F(y)$ or $$\frac{(1-t^2)^{3/2}}{t^3}F(y)\tag{2}$$ Since $t\to 0^+$ the desired limit is the same as that of $F(y) /t^3$.
Using Taylor series we can write $F(y) $ as $$2\left(\frac{1}{t^2}-\frac{1}{3}\right)\left(t+\frac{t^3}{3}+\frac{t^5}{5}+o(t^5)\right)-\frac{2}{t}$$ And this can be written as $$2\left(\frac{1}{t}+\frac{t}{3}+\frac{t^3}{5}-\frac{t}{3}-\frac{t^3}{9}+o(t^3)-\frac{1}{t}\right)=\frac{8t^3}{45}+o(t^3) $$ It follows that $F(y) /t^3$ tends to $8/45$.
One can observe that the substitution $1+y=1/t^2$ helps here a lot as it simplifies the log term to a well known Taylor series.
A: To calculate the limit for $y\to\infty$ you can use L'Hospital, but not immediately as the term will be $\infty \cdot 0 - \infty$ which is not defined. You first have to convert this term to get an expression of $0/0$ so that L'Hospital can really be used.
To begin with, the functions are defined with $F,G,H$ so that the above formula is simplified to
$$
F\cdot G - H
$$
with $F\to \infty, G \to 0, H \to \infty$.
Transforming this into a formula that is allowed for L'Hospital yields
$$D = \frac{1}{\frac{1}{F\cdot H}} \cdot \left(G\cdot \frac 1H - \frac 1F\right)$$
as for $y\to\infty$ it would approach $0/0$
Let's calculate each function and its derivative for its own in terms of maximum exponents - as that is the only interesting at the end when looking at the behavior for $y\to\infty$. The behavior for large $y$ is denoted with $\sim$ which is similar to $\Theta$ in the $\mathcal{O}$ notation:
\begin{align}
F &\sim y^1 \\
G &\sim 0 \\
H &\sim y^{- \frac 12} \\
\frac{d}{dy} G &= \frac{\sqrt{1+y}-1}{\sqrt{1+y}+1}\cdot\frac{\frac{1}{2\sqrt{1+y}}(\sqrt{1+y}-1) - \frac{1}{2\sqrt{1+y}} (\sqrt{1+y}+1)}{(\sqrt{1+y}-1)^2}  \\
&= \frac{-1}{(\sqrt{1+y}+1)(\sqrt{1+y}-1)(\sqrt{1+y})} \sim y^{-\frac{3}{2}}\\
\frac{d}{dy} \frac{1}{H} &=\frac{-1}{4(1+y)^{\frac{3}{2}}} \sim y^{-\frac 32} \\
\frac{d}{dy}\frac{1}{F} &= \frac{1}{-\left(y+\frac 23\right)^{2}} \sim y^{-2} \\
\frac{d}{dy}G\frac{1}{H} &\sim y^{-\frac 32} \cdot y^{-\frac 12} + 0 \cdot y^{-\frac 32} \sim y^{-2} \\
\frac{d}{dy}\frac{1}{F}\frac{1}{H} &\sim y^{-2} y^{-\frac 12} + y^1 y^{-\frac32} \sim y^{-\frac 52} + y^{-\frac 12} \sim y^{-\frac 12}
\end{align}
If we put everything together now, we conclude:
\begin{align}
D \sim \frac{1}{y^{-\frac 12}} \cdot (y^{-2} - y^{-2}) \sim y^{\frac 12} \cdot y^{-2} \sim y^{-\frac 32}\\
\end{align}
It could be possible that we get $y^{-2}-y^{-2}=0$ in the brackets, which would change the result. But if we take a look at the coefficient in front of $y^{-2}$ we get $(-1)\cdot \frac 12 = -\frac 12$ for $\frac{d}{dy}\left(G\cdot \frac 1H\right)$ (the factor of $\left(\frac{d}{dy}G\right)\frac{1}{H}$ from using the product rule as the other product has exponent $0$) and $-1$ for $\frac{d}{dy}\frac 1F$. As $-\frac 12 \neq -1$ the above difference will not be zero, therefore the exponents will remain as calculated.
So in total we have $D\sim y^{-\frac 32}$ or more mathematically:
$$
D \in \Theta\left(y^{-\frac 32}\right)
$$
Puh. That took a long time to write and think about. I hope everything is correct and clearly written. If anything is unclear or any questions arise, please ask. I would appreciate any feedback in form of comments or votes - especially as this is my first "big" answer with a more difficult topic :)
A: Set $x=\sqrt{1+y}-1$. Then $y=(x+1)^2-1=x^2+2x$, and your expression simplifies to
$$ \bigl(x^2+2x+\tfrac23 \bigr)\log\bigl(1+\tfrac2x \bigr) - 2x - 2 $$
Expanding the logarithm in powers of $2/x$ gives us
$$ \bigl(x^2+2x+\tfrac23 \bigr)\Bigl(\frac2x-\frac2{x^2}+\frac{8}{3x^3}-\frac{4}{x^4}+\frac{32}{5x^5}+o(x^{-5}) \Bigr) - 2x - 2 $$
We can then multiply out and collect terms to get
$$ \begin{align} (2-2)&x^1 + {}
\\ (-2 + 4 - 2)&x^0 +{}
\\ (\tfrac83 -4 + \tfrac43)&x^{-1} +{}
\\ (-4+\tfrac{16}3 - \tfrac43)&x^{-2} +{}
\\ (\tfrac{32}5 -8 + \tfrac{16}{9})&x^{-3} + o(x^{-3})
= \tfrac{8}{45}x^{-3} + o(x^{-3})
\end{align}
$$
Since asymptotically $x\sim y^{1/2}$ this is also
$$\frac{8}{45}y^{-3/2} + o(y^{-3/2}).$$
A: Considering $$A=\left(y+\frac23\right)\log\left(\dfrac{\sqrt{1+y}+1}{\sqrt{1+y}-1} \right) - 2 \sqrt{1+y}$$ first simplify
$$A=\left(y+\frac{2}{3}\right) \log \left(\frac{y+2 \sqrt{y+1}+2}{y}\right)-2
   \sqrt{y+1}$$ For large values of $y$, you could start using $y=\frac 1x$ to make
$$A=\frac{(2 x+3) \log \left(2 x+2 \sqrt{x (x+1)}+1\right)-6 \sqrt{x (x+1)}}{3 x}$$ Now, Taylor series
$$\sqrt{x (x+1)}=x^{1/2}+\frac{x^{3/2}}{2}-\frac{x^{5/2}}{8}+\frac{x^{7/2}}{16}+O\left(x^{9/2}\right)$$
$$ \log \left(2 x+2 \sqrt{x (x+1)}+1\right)=2 x^{1/2}-\frac{x^{3/2}}{3}+\frac{3 x^{5/2}}{20}-\frac{5 x^{7/2}}{56}+O\left(x^{9/2}\right)$$
$$A=\frac{8 x^{3/2}}{45}-\frac{4 x^{5/2}}{35}+O\left(x^{7/2}\right)$$ Back to $y=\frac 1x$ gives the result.
As I used to say : we are always closer to $0$ than to $\infty$
