What are some special tricks on investigating limits? During studying limits I regularly come across some of them, that cannot be simplified anymore using standard laws and formulas (e.g. substitutions, sum/product rules, L'Hôpital, etc.). However when the final expression is received it is still not obvious, how the function (resp. series) converges at a certain limit. 
For instance
$$
    \lim_{s \to 0} {e^s-1 \over s} = 1 \;\;, \;\; \lim_{x \to \infty}e^{\log x \over x} = 1
$$ 
I somehow instinctively understand, that I should pay attention to how functions behave, which of them grows (resp. falls) faster than another. However I would like to know if there're some solid methods of analyzing such cases.
 A: Yes, use the fact that $$e^s = 1+ s + O(s^2).$$
Then we get $$\frac{e^s - 1}{s} = \frac{s + O(s^2)}{s} \to 1.$$
Alternatively note that the limit is equivalent to computing,  
$$ \lim_{x \to 0} \frac{e^s - 1}{s}  =  \lim_{x \to 0} \frac{e^s - e^0}{s - 0} = \left (e^s)'\right|_{s=0} = e^0 = 1.$$
A: If you're willing to accept that $\sum_{n=1}^\infty \frac{x^{n-1}}{n!}$ is uniformly convergent then you can use this to evaluate the first limit
$$
\lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{x \to 0} \sum_{n = 1}^\infty \frac{x^{n-1}}{n!} = \sum_{n=1}^\infty \frac{0^{n-1}}{n!} = 1
$$
In general for problems like these you can use power series to evaluate the limits. In other words if you know the power series representation of a function you can plug it into the limit and do some manipulation there.
For the second series I would take advantage of the continuity of $e^x$ and prove that
$$
\lim_{x \to \infty} x\sqrt[x]{e} = 1
$$
which we can see
$$
\lim_{x \to \infty} x\sqrt[x]{e} = \lim_{x \to \infty} \frac{x}{e^x} = \lim_{x \to \infty} \frac{x}{\sum_{n=0}^\infty \frac{x^n}{n!}} = \lim_{x \to \infty} \frac{1}{\frac{1}{x} + \sum_{n=1}^\infty \frac{x^{n-1}}{n!}} = \lim_{x \to \infty} \frac{1}{1 + \frac{1}{x} + \sum_{n=2}^\infty \frac{x^{n-1}}{n!}} = 1
$$
Generally I like looking at power series to solve these limits.
Just thought of the somewhat traditional way to prove your first limit. Define $e = \lim_{n \to \infty} (1 + 1/n)^{n}$ and notice
$$
\lim_{n \to \infty} (1 + 1/n)^n = \lim_{h \to 0} (1 + h)^{1/h}
$$
so then
$$
\lim_{s \to 0} \frac{e^s - 1}{s} = \lim_{s \to 0} \frac{\lim_{h \to 0} (1+h)^{s/h}-1}{s}
$$
Now take advantage of $e^x$'s continuity to combine the limits
$$
\lim_{s \to 0} \frac{(1+s)^{s/s}-1}{s} = 1
$$
A: I think there aren't any methods that would cover every case,for example in this case $e^0=1$,and by the continuity of the exponential function as $s\to 0,e^s\to 1$ so from that $e^s-1\to 0$ and by the limit $s\to 0$ so it's 1.Also the above method with $O(s)$ is useful.Now for the second limit $x>\log(x)$ or $e^x>x$ since $e^x$ grows much faster than $x$  we can say that $\frac{\log(x)}{x}\to0$ as $x\to\infty$ so again since $e$ is continuos we say that limit is $1$
