Is it possible to evaluate $\lim_{x\to\infty}\frac{\ln(x+e^x+e^{2x})}{x}$ without applying L'Hopital rule? $$\lim_{x\to\infty}\frac{\ln(x+e^x+e^{2x})}{x}$$
In the book I am reading, evaluated the limit with L'Hopital rule like this:
$$\lim_{x\to\infty}\frac{\ln(x+e^x+e^{2x})}{x}=\frac{\infty}{\infty}\rightarrow \lim_{x\to\infty}\frac{1+e^x+2e^{2x}}{x+e^x+e^{2x}} $$
Then it used equivalence and write $\lim_{x\to\infty}\cfrac{2e^{2x}}{e^2x}=2$.
My approach to evaluate $\lim_{x\to\infty}\frac{\ln(x+e^x+e^{2x})}{x}$ is different:
in the numerator of the fraction and inside the $\ln()$ the function $e^{2x}$ goes faster to infinity than $x$ , $e^x$. therefore we can ignore these two:
$$\lim_{x\to\infty}\frac{\ln(e^{2x})}{x}=\lim_{x\to\infty}\frac{2x}{x}=2$$
Is my approach right? and was it really necessary to use L'Hopital Rule?
 A: Your approach is correct. For a more rigorous argument do the following:
$\frac {\ln (x+e^{x}+e^{2x})} x=\frac {\ln [e^{2x} (xe^{-2x}+e^{-x}+1)} x=\frac {2x+\ln (xe^{-2x}+e^{-x}+1)} x$. Using the fact that $\ln (1+t) \sim t$ as $t \to 0$ we get $\lim \frac {2x+xe^{-2x}+e^{-x}} x=2$.
A: No, it is not really necessary to use L'Hopital's rule. You can do it as follows:\begin{align}\lim_{x\to\infty}\frac{\log(x+e^x+e^{2x})}x&=\lim_{x\to\infty}\frac{\log\left(e^{2x}\left(\frac x{e^{2x}}+\frac1{e^x}+1\right)\right)}x\\&=\lim_{x\to\infty}\frac{2x}x+\frac{\frac x{e^{2x}}+\frac1{e^x}+1}x\\&=2+0=2.\end{align}On the other hand, although your ideae is fine, I think that your justification is far from complete.
A: You are right thinking of $e^{2x}$ as the term that's going to diverge faster, and the idea of using it is correct. However, if you want a somewhat more rigorous proof I'd recommend the following steps:
$\displaystyle\lim_{x\to+\infty}\dfrac{\ln(e^{2x}+e^x+x)}{x}=\lim_{x\to+\infty}\dfrac{\ln(e^{2x}(1+\frac{1}{e^x}+\frac{x}{e^{2x}}))}{x}=\lim_{x\to+\infty}\dfrac{\ln(e^{2x})+\ln(1+\frac{1}{e^x}+\frac{x}{e^{2x}})}{x}=\lim_{x\to+\infty}\dfrac{\ln(e^{2x})}{x}=\lim_{x\to+\infty}\dfrac{2x}{x}=2$
A: We have that for any $\varepsilon>0$ eventually
$$2=\frac{\ln(e^{2x})}{x}\le \frac{\ln(x+e^x+e^{2x})}{x}\le\frac{\ln(e^{(2+\varepsilon)x})}{x}=2+\varepsilon$$
therefore since $\varepsilon$ is arbitrarily small
$$\frac{\ln(x+e^x+e^{2x})}{x} \to 2$$
