help on limit exercise I'm having some trouble solving this limit: 
$$\lim_{x\to\infty} x\left[\left(\cosh x\right)^ \frac1x - \left(1+\frac1x\right)^x\right]$$
It's part of a set of limits wich should be solved using taylor.
I tried this road: 
$$\lim_{x\to\infty} x\left[e^{\frac1x\ln(\cosh x)}-e^{\ln\left(1+\frac1x\right)x}\right]$$
I then tried with algebraic manipulation using the definition of hyperbolic cosine $\frac12(e^x+e^{-x})$ and then I also played a bit with l'Hopital but it turns into something suspiciously compicated...
I'm taking real analysis 1 and my toolset is:
-algebraic manipulation
-talyor series
-l'Hopital (if and only if all else fails)
I can't use more advanced techniques since they are not part of the course. I'm sure there's something obvious I'm missing.
Any idea on how to proceed?
 A: I have tried several approaches, and the fastest seems the following [it does not require L'Hopital]:
$$
\lim_{x\to\infty} x[e^{\ln(\cosh x)\frac1x}-e^{\ln(1+\frac1x)x}]=\lim_{x\to\infty} e^{\ln x}[e^{\ln(\cosh x)\frac1x}-e^{\ln(1+\frac1x)x}]
$$
$$
\lim_{x\to\infty} e^{\phi_1(x)}-e^{\phi_2(x)}\ .
$$
Let us analyze the series expansions of the two functions in the exponent up to second next-to-leading order
$$
\phi_1(x)=\ln x+\frac{1}{x}\ln(\cosh(x))=\ln x+\frac{1}{x}\ln\left(\frac{1}{2}(e^x+e^{-x})\right)=\ln x-\frac{\ln 2}{x}+1+\frac{1}{x}\ln(1+e^{-2x})
$$
$$
\sim \ln x+1-\frac{\ln 2}{x}
$$
$$
\phi_2(x)=\ln x+x \ln(1+1/x)\sim\ln x +1-\frac{1}{2x}
$$
Therefore, Taylor-expanding the exponentials [of the type $e^{-c/x}$] around $x=\infty$ 
$$
e^{\phi_1(x)}\sim e x e^{-\ln 2/x}\sim e x -e \ln (2)+\frac{e \ln ^2(2)}{2 x}-\frac{e \ln ^3(2)}{6 x^2}+\ldots
$$
$$
e^{\phi_2(x)}\sim e x-\frac{e}{2}+\frac{e}{8 x}-\frac{e}{48 x^2}+\ldots
$$
we get the final result by subtracting one from the other
$$
\boxed{\lim_{x\to\infty} x[e^{\ln(\cosh x)\frac1x}-e^{\ln(1+\frac1x)x}]=\frac{e}{2}-e\ln 2\approx -0.525028...}
$$
