What am I doing wrong finding $\lim_{x\to\infty}\frac{\ln (2+ e^{3x})}{3+e^{2x}}$? $$\lim_{x\to\infty}\frac{\ln (2+ e^{3x})}{3+e^{2x}}$$
$$\lim_{x\to\infty}\frac{3\cdot e^{3x}\cdot \frac{1}{2+e^{3x}}}{2\cdot e^{2x}}$$
$$\frac{3}{2}\lim_{x\to\infty}{\frac{e^x}{2+e^{3x}}}$$
$$\frac{3}{2}\lim_{x\to\infty}{\frac{1}{\frac{2}{e^x}+e^{2x}}}$$
$$\frac{3}{2}\cdot \frac{1}{0+\infty}$$
$$0$$
The answer in the book is $\frac{3}{2}$. Taylor's expansion also leads me to $0$. Is it my mistake, or is there a book typo?
 A: I tried it out myself. The answer is indeed zero and it is a book typo.
p.s. Wolfram: https://www.wolframalpha.com/input/?i=limit+%5Cfrac%7B%5Cln+(2%2B+e%5E%7B3x%7D)%7D%7B3%2Be%5E%7B2x%7D%7D
A: *

*there is probably a typo in the book, if answer is $\frac 32$ then we miss a log on the denominator

*your second line is wrong

*avoid carrying the limit operator, use equivalents instead.


$\dfrac{\ln(2+e^{3x})}{\ln(3+e^{2x})}\sim\dfrac{\ln(e^{3x})}{\ln(e^{2x})}\sim\dfrac{3x}{2x}\to \dfrac 32$
With the typo:
$\dfrac{\ln(2+e^{3x})}{3+e^{2x}}\sim\dfrac{\ln(e^{3x})}{e^{2x}}\sim\dfrac{3x}{e^{2x}}\to 0$
And if you do not want to use equivalents, just factor the dominant terms until you get something whose limit is easy to determine:
$\dfrac{\ln(2+e^{3x})}{3+e^{2x}}=\dfrac{\ln(e^{3x}(1+2e^{-3x}))}{e^{2x}(1+3e^{-2x})}=\dfrac{\ln(e^{3x})+\ln(1+2e^{-3x})}{e^{2x}(1+3e^{-2x})}=\dfrac{3x+\ln(1+2e^{-3x})}{e^{2x}(1+3e^{-2x})} = $
$$\underbrace{3xe^{-2x}}_{\to 0}\underbrace{\left(\dfrac{1+\overbrace{\dfrac{\ln(1+2e^{-3x})}{3x}}^{\to 0}}{1+\underbrace{3e^{-2x}}_{\to 0}}\right)}_{\to 1}\to 0$$
similarly as above.
