Evaluate $\lim_{x\to +\infty}\frac{x(\sqrt{x^2+x}-x) + \cos x\ln x}{\ln (1+\cosh x)}$ 
$$\lim_{x\to +\infty}\frac{x(\sqrt{x^2+x}-x) + \cos x\ln x}{\ln (1+\cosh x)}$$

I'm trying to apply the Squeeze Theorem, but I'm having troubles evaluating the lower and upper sides of the inequality.
My (poor) attempt (for readiness I'll drop the $\lim_{x\to +\infty}$ part). 
$$\frac{x(\sqrt{x^2+x}-x) + \cos x}{\ln (1+\cosh x)} \leq \frac{x(\sqrt{x^2+x}-x) + \cos x\ln x}{\ln (1+\cosh x)} \leq \frac{x(\sqrt{x^2+x}-x) + \ln x}{\ln (1+\cosh x)}$$
Considering that as $x \to +\infty$:
$$\sqrt{x^2+x} \approx \sqrt{x^2} = x$$
and
$$\ln x = o(x)$$
we have that
$$\frac{\cos x}{\ln (1+\cosh x)} \leq \frac{x(\sqrt{x^2+x}-x) + \cos x\ln x}{\ln (1+\cosh x)} \leq \frac{ x}{\ln (1+\cosh x)}$$
From I don't know how to proceed. I'm not sure whether I can take a Taylor Series if as $x \to +\infty$ but in any case It doesn't seem to me that I haven't gone to far.
 A: I think you are making this a bit too hard given you are already using Taylor series (technically these are Laurent series, but they work the same way and are easy enough to calculate). First note that, for $x$ about infinity, we can write
$$x+\frac{1}{2}\sim \sqrt{x^2+x}$$
Proving this just requires dividing one side by the other and taking a limit.
Next, note that, about infinity, $$\ln(1+\cosh x)\sim x-\ln(2)$$ 
Where this last asymptotic follows from writing $\cosh x$ in it's exponential form and simplifying the logarithm, remembering that $\ln(e^x+1)\sim x$. We now rewrite your limit as such: 
$$\lim_{x\to \infty}\frac{\frac{x}{2} + \cos x\ln x}{x-\ln(2)}=\lim_{x\to \infty}\frac{\frac{1}{2} + \frac{1}{x}\cos x\ln x}{1-\frac{1}{x}\ln(2)}$$
Now all that is necessary to note is that $\frac{\cos x \ln x}{x}$ is bounded by $\frac{-\ln x}{x}$ and $\frac{\ln x}{x}$, which both trivially go to zero. As such, we solve your original limit in just one step. Here is some graphical "proof" if you want it! The red curve represents the original function, and the blue curve represents the simpler asymptotic function

A: The left part of inequality is wrong. (Because $-1 \le \cos x \le 1$)
$$\frac{x(\sqrt{x^2+x}-x) - \ln x}{\ln (1+\cosh x)} \leq \frac{x(\sqrt{x^2+x}-x) + \cos x\ln x}{\ln (1+\cosh x)} \leq \frac{x(\sqrt{x^2+x}-x) + \ln x}{\ln (1+\cosh x)}$$
Finding the right part :-
$$\lim_{x\to +\infty}\frac{\ln x}{\ln (1+\cosh x)} =  \lim_{x\to +\infty}\frac{\ln x}{\ln e^x}\frac{\ln e^x}{\ln (1+\frac{e^x+e^{-x}}{2})} =  \lim_{x\to +\infty}\frac{\ln x}{x}\times1=0$$
Finding the left part. Here's the trick:-
$$(\sqrt{x^2+x}-x) = \frac{(\sqrt{x^2+x}-x)(\sqrt{x^2+x}+x)}{(\sqrt{x^2+x}+x)}=\frac{x}{(\sqrt{x^2+x}+x)}$$
$$\lim_{x\to +\infty}\frac{x}{(\sqrt{x^2+x}+x)}= \lim_{x\to +\infty}\frac{1}{(\sqrt{1+\frac{1}{x}}+1)}=\frac{1}{2}$$
The remaining part:-
$$\lim_{x\to +\infty}\frac{ x}{\ln (1+\cosh x)} =  \lim_{x\to +\infty}\frac{ x}{\ln e^x}\frac{\ln e^x}{\ln (1+\frac{e^x+e^{-x}}{2})} =  1\times1=1$$
Therefore limit is $1\times\frac{1}{2}+0=\frac{1}{2}$
