I have the following limit:
$$\lim_{x\rightarrow+\infty}\frac{\log\left({e^{2x+1}}+x\right)}{x}=2$$
I know for sure that
$$\lim_{x\rightarrow+\infty}\frac{\log{x}}{x}=0$$
Because $x$ grows a lot faster than $\log{x}$. Then why wouldn't the first limit be equal to $0$ as well?
because it has that $e^{2x+1}$ in the numerator it is making all the difference
$$\lim_{x\rightarrow+\infty}\frac{\log\left({e^{2x+1}}\right)}{x}\le\lim_{x\rightarrow+\infty}\frac{\log\left({e^{2x+1}}+x\right)}{x}$$
$$\lim_{x\rightarrow+\infty}\frac{\left({{2x+1}}\right)}{x}\le\lim_{x\rightarrow+\infty}\frac{\log\left({e^{2x+1}}+x\right)}{x}$$ $$2\le\lim_{x\rightarrow+\infty}\frac{\log\left({e^{2x+1}}+x\right)}{x}$$
find the limit using Lhopital's rule
$$\lim_{x\rightarrow+\infty}\frac{\log\left({e^{2x+1}}+x\right)}{x}=\lim_{x \to 0}\frac{e^{2x+1 }\cdot 2+1}{e^{2x+1} +x}$$
$$\lim_{x\rightarrow+\infty}\frac{\log\left({e^{2x+1}}+x\right)}{x}=\lim_{x \to 0}\frac{2+\frac{1}{e^{2x+1}}}{1+\frac{x}{e^{2x+1}}}=2 $$