I am currently doing some old math exams and was wondering whether my solution for the given question was appropriate (see picture)
for me it just feels like the point where I put together both limits bc the functions are continous I am doing something wrong, especially because the problem explicitly does not have a limit in front of the summand given by β*e^x+γ
Really appreciate your help guys, and sorry I am not yet able to write in LaTex.
$$\lim_{x\to+\infty}\left(\frac{2e^{3x}}{e^{2x}-1}-\beta e^x +\gamma \right)=0 $$ Find the common denominator of your expression: $$\frac{2e^{3x}-\beta e^{3x}+\beta e^x+\gamma e^{2x}-\gamma}{e^{2x}-1} $$ If $2-\beta\neq0$, the limit it $+\infty$ since $e^{3x}$ grows faster than $e^{2x}$; so you must have $\beta=2$.
If $\gamma\neq0$, the limit is $\gamma$ , hence you must have $\gamma=0.$