To calculate $$\lim\limits_{x\to 0^{+}}x\ln\left(x\sinh(\dfrac{1}{x}) \right) $$
the author of textbook use the followings method which there is some steps i would like to understand
$x\sinh\left(\frac1x\right)\sim_{0^+} \frac{x}2e^{\frac1x}\underset{x\to 0^{+}}{\longrightarrow }+\infty\neq 1$
Then $$\ln\left(x\sinh(\dfrac{1}{x}) \right)\sim_{x\to 0^+}\ln\left( \dfrac{x}{2}e^{\dfrac{1}{x}} \right)=\dfrac{1}{x}+\ln\left( \dfrac{1}{x}\right)\sim_{x\to 0^{+}}\dfrac{1}{x}$$
Thus $$\lim\limits_{x\to 0^{+}}x\ln\left(x\sinh(\dfrac{1}{x}) \right)=1 $$
here's steps that i would like to understand :
- $x\sinh\left(\frac1x\right)\sim_{0^+} \frac{x}2e^{\frac1x}$
- $\dfrac{1}{x}+\ln\left( \dfrac{1}{x}\right)\sim_{x\to 0^{+}}\dfrac{1}{x}$
what i know :
$$ x\sinh(x)=x\dfrac{e^{x}-e^{-x}}{2}=\dfrac{x}{2}\left(e^{x}-e^{-x} \right) $$
To easy understanding let's have $u=\frac 1x$ so $\lim\limits_{x\to 0^+}u=+\infty$.
$\displaystyle x\sinh(\frac 1x)=\frac{\sinh(u)}u=\frac{e^u-e^{-u}}{2u}=\frac{e^u}{2u}(1-\underbrace{e^{-2u}}_{\to 0})$
When $u\to+\infty$ then $e^u\to+\infty$ and $e^{-u}\to 0$ so this term is negligible which I showed by factorizing the main contribution term.
And we have $\displaystyle x\sinh(\frac 1x)\sim\frac{e^u}{2u}=\frac x2e^{\frac 1x}$
Similarly
$\displaystyle \frac 1x+\ln(\frac 1x)=u+\ln(u)=u(1+\underbrace{\frac{\ln(u)}u}_{\to 0})$
When $u\to+\infty$ then $\ln(u)$ is negligible in regard to $u$ which I showed by factorizing the main contribution term.
And we have
$\displaystyle \frac 1x+\ln(\frac 1x)\sim u=\frac 1x$
But I do not see the utility of this particular step in order to solve the current problem.
Overall I would do instead:
$x\ln(x\sinh(\frac 1x))\sim x\ln(\frac x2e^{1/x})=x\big(\ln(x)-\ln(2)+\ln(e^{1/x})\big)=\underbrace{x\ln(x)}_{\to 0}+\underbrace{x\ln(2)}_{\to 0}+\underbrace{\frac xx}_{\to 1}\to 1$