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?

HINT Use $\sinh^{-1}(x) = \ln (x + \sqrt{1+x^2})$.

 $$\displaystyle \lim_{x \rightarrow \infty} \dfrac{\ln(x^2+4)}{\ln(x+\sqrt{1+x^2})} = \displaystyle \lim_{x \rightarrow \infty} \dfrac{\ln(x^2) + \ln(1+4/x^2)}{\ln(x) + \ln(1+\sqrt{1+1/x^2})}$$
 $$ = \displaystyle \lim_{x \rightarrow \infty} \dfrac{\ln(x^2)}{\ln(x)} \dfrac{\left(1+\dfrac{\ln(1+4/x^2)}{\ln(x^2)} \right)}{\left(1 + \dfrac{\ln(1+\sqrt{1+1/x^2})}{\ln(x)} \right)}$$
 $$ = \displaystyle \lim_{x \rightarrow \infty} 2 \dfrac{\ln(x)}{\ln(x)} \lim_{x \rightarrow \infty} \dfrac{\left(1+\dfrac{\ln(1+4/x^2)}{\ln(x^2)} \right)}{\left(1 + \dfrac{\ln(1+\sqrt{1+1/x^2})}{\ln(x)} \right)}$$
 $$ = 2 \displaystyle \lim_{x \rightarrow \infty} \dfrac{\left(1+\dfrac{\ln(1+4/x^2)}{\ln(x^2)} \right)}{\left(1 + \dfrac{\ln(1+\sqrt{1+1/x^2})}{\ln(x)} \right)} =2 $$

