How to deal with $g(f(x))$ in the limit? Let $\lim\limits_{x \to \infty} f(x)=\infty$. Does the following hold as $x\to\infty$?:
$$\exp(x) \leq \exp\left({\frac{x}{1+1/\ln(1+f(x))}}\right)$$
My effort: If above inequality holds in the limit, the it must be that the following holds in the limit (take $\log$ of both sides): 
$$x \leq {\frac{x}{1+1/\ln(1+f(x))}}$$
Or, equivalenty:
$$1/{\ln(1+f(x))} \leq 0$$
which is valid since $f(x)\to \infty$ (given in the problem statement), and thus $1/{\ln(1+f(x))} \to 0$. So, the inequality in the problem statement holds in the limit.
Is this thought process correct? If not, please provide me with a correct proof if the inequality above holds or does not hold for all $f(x)$ that tend to infinity as $x \to \infty$.
 A: So the problem with your reasoning comes when you claim
$$
1/\log(1+f(x)) \le 0
$$
Since $f(x) \rightarrow \infty$, this means that $f(x) > 0$ for all $x$ sufficiently large. When $f(x) > 0$ however, we would have that $1/\log(1+f(x)) >0$ because $\log(z) > 0$ whenever $z > 1$.
All the other steps in your reasoning are correct, so in fact the answer is that the inequality is false for large $x$.
The point you seem to be confused on is the fact that 
$$
\lim_{x\rightarrow\infty} 1/\log(1 + f(x)) \le 0
$$
which is true, but is not good enough to make the original inequality true. Why? Because both sides go to infinity, so even though the $1/\log(1 + f(x))$ is trying to go to $0$, since it is strictly positive it can still affect the rate at which the function $\exp(\frac{x}{1 + 1/\log(1 + f(x))})$. For example, suppose $f(x) = x - 1$. Clearly, that goes to infinity. You would therefore be claiming that in the limit
$$
\exp(x) \le \exp\left(\frac{x}{1 + 1/\log(1+f(x))}\right) = \exp\left(\frac{x}{1 + 1/\log(x)}\right) = \exp\left(\frac{x\log(x)}{\log(x) + 1}\right)
$$
but the left hand side goes to infinity much faster than the left. If we take the ratio of both sides, you would claim:
\begin{eqnarray}
\frac{\exp(x)}{\exp\left(\frac{x\log(x)}{\log(x) + 1}\right)} &\le& 1 \\
\exp\left(x - \frac{x\log(x)}{\log(x) + 1}\right) &\le& 1 \\
\exp\left(x\left(1 - \frac{\log(x)}{\log(x) + 1}\right)\right) &\le& 1 \\
\exp\left(x\left(\frac{1}{\log(x) + 1}\right)\right) &\le& 1
\end{eqnarray}
but the right hand side goes to infinity in the limit.
A: We have that, since exp function is increasing
$$e^x \leq e^{\left({\frac{x}{1+1/\ln(1+f(x))}}\right)}\iff x \leq {\frac{x}{1+1/\ln(1+f(x))}}$$
and since eventually $x>0$, $1+f(x)>1$ we have
$$x \leq {\frac{x}{1+1/\ln(1+f(x))}} \iff \frac1{\ln(1+f(x))} \le 0$$
which is not true.
A: Your thought process is rather tricky, as your last line holds exactly in the limit of $x\to\infty$, and is wrong for all other $x>0$.
You however start your deduction from the desired result and move backwards - which is okay, as every step you make is bijective for sufficiently big $x$, but to make conclusions about the bijectivity of your steps regarding a limit, you need to prove that the limit exists (remember, all those rules one learns require this).
This is not given in your proof, as e.g. $lim_{x\to\infty} exp(x) = \infty$.
A: Since all of this process is reversible so your method is right but we have $$\dfrac{1}{\ln{1+f(x)}}\le 0\leftarrow\rightarrow\ln{1+f(x)}< 0$$which means that $$f(x)<0$$so it doesn't hold under your constraints.
