Approximation of function when variable tends to infinity

Question

I'm reading these notes and on page 99 they approximate the function $$ q_2(x,t) = -12 \frac{3 + 4 \cosh(2x+24t) + \cosh(4x)}{\left( 3 \cosh(x-12t)+\cosh(3x+12t) \right)^2} $$ for when $t \rightarrow \infty$ as follows $$ q_2(x,t) \sim -12 \frac{2e^{2x+24t}}{\left( \frac{3}{2} e^{12t-x} + \frac{1}{2}e^{3x+12t} \right)^2} = \frac{-8}{\cosh^2 \left(2x-\frac{1}{2} \ln(3) \right)} $$ I don't understand neither of the two steps and any help would be much appreciated.

Answer 1

$$\cosh(x)= \frac{e^x+e^{-x}}{2}$$ $$q_2(x,t)= -12 \frac{3+4\cosh(2x+24t)+\cosh(4x)}{(3\cosh(x-12t)+\cosh(3x+12t))^2}$$ Substituting in we have $$= -12 \frac{3+2(e^{(2x+24t)}+e^{-(2x+24t)})+\frac{e^{4x}+e^{-4x}}{2}}{(\frac{3}{2}e^{(x-12t)}+e^{-(x-12t)}+\frac{e^{3x+12t}+e^{-(3x+12t)}}{2})^2}$$ Taking the limit as $t \rightarrow \infty$ means that $e^{-t} \rightarrow 0$. This simplifies the above to $$= -12 \frac{3+2e^{2x+24t}+\frac{e^{4x}+e^{-4x}}{2}}{(\frac{3}{2}e^{-(x-12t)}+\frac{e^{3x+12t}}{2})^2}$$ This is the first way that it is represented. It will take a little more work but then you can re-write it using the same definition of $\cosh(x)$ to write it in the alternate form.