Calculate difficult limits I want to improve my counting limits. I've found some difficult examples:


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*$\displaystyle\lim_{x \to +\infty}\left((x+1)^{1+\frac1x}-x^{1+\frac{1}{x+a}}\right)$

*$\displaystyle\lim_{x\to +\infty}x^2(\arctan x - \frac{\pi}{2})+x$

*$\displaystyle\lim_{x\to+\infty}\left( \sqrt[3]{x^3+x^2+x+1}-\sqrt{x^2+x+1}\cdot\frac{\ln(e^x+x)}{x} \right)$

*$\displaystyle\lim_{x\to 0} \left(\frac{a^x-x\ln a}{b^x-x\ln b} \right)^{\frac{1}{x^2}}$
and I don't know how to touch them. I know: L'Hôpital's rule, Mean value theorem, Taylor's theorem but still don't have this skill. Can anybody help me?
 A: Usually series expansions are the best way to go.  L'Hospital should be avoided whenever possible.
For example in 1)
$$ (x+1)^{1+1/x} = \exp\left((1+\frac1x) \ln(x+1)\right) = \exp\left((1+\frac1x)(\ln(x)+\ln(1+\frac1x))\right)$$
$$ = \exp\left(\ln(x) + \frac{\ln(x)}{x} + \frac1x + O(\frac{1}{x^2})\right) = x \left(1 + \frac{\ln(x)}{x} + \frac{1}{x} + O(\frac{\ln(x)^2}{x^2})\right)$$
$$ = x + \ln(x) + 1 + o(1)$$
while
$$ x^{1+1/(x+a)} = x \exp\left(\frac{\ln(x)}{x+a}\right) = x \exp\left(\frac{\ln(x)}{x} + O(\frac{\ln (x)}{x^2})\right)$$
$$ = x \left(1 + \frac{\ln(x)}{x} + O(\frac{\ln(x)^2}{x^2})\right)$$
$$ = x + \ln(x) + o(1)$$
so the limit of their difference is $1$.
EDIT: for 2) note that $$\arctan(t) = \frac{\pi}{2} - \arctan(1/t) =  \frac{\pi}{2} - \frac{1}{t} + O(\frac{1}{t^3})$$
A: (2) Rewrite it as $$\lim_{x\to\infty}\frac{x\arctan x-\frac{\pi}2 x+1}{\frac1x}$$ and apply l’Hospital’s rule twice, followed by a little algebraic simplification.
(4) This appears to succumb to the usual technique for such problems: let $L$ be the desired limit, take logs to get 
$$\ln L=\lim_{x\to 0}\frac{\ln(a^x-x\ln a)-\ln(b^x-x\ln b)}{x^2}\;,$$
and beat it to death with l’Hospital’s rule.
