Evaluating $\lim\limits_{x\to\infty} \frac{\ln(x^2+4)}{\sinh^{-1}x}$ 
$$\lim\limits_{x\to\infty} \frac{\ln(x^2+4)}{\sinh^{-1}x}$$

This is an exam practice question. BTW, I am refering to the inverse hyperbolic function above.
Since this is infinity/infinity, I used one application of L'Hopital's rule for this one, and then did a bit of algebraic manipulation to get $\frac{2}{1}$, i.e. $2$. Is this right?
Is there a quicker way to do this? It seems like Taylor Polynomials are often used, but they confuse me - Is this one, where Taylor Polynomials would have been easier?

Another possibility: substitution $\sinh^{-1}(x)=t$.
You have: 


*

*$t\to\infty$ for $x\to\infty$

*$x=\sinh t= \frac{e^t-e^{-t}}2$

*$x^2+4=\frac{e^{2t}+e^{-2t}}4+3$


So your limit becomes
$$\lim\limits_{t\to\infty} \frac{\ln\left(\frac{e^{2t}+e^{-2t}+12}4\right)}t=
\lim\limits_{t\to\infty} \frac{\ln(e^{2t}+e^{-2t}+12)-\ln 4}t,$$
which should not be that difficult.
