Explanation for $\lim_{x\to2} e^{\frac{1}{x-2}}$ I can't find out why is the limit from the left side = 0 and from the right = Infinity? 

First look at $\dfrac1{x-2}$. It has a vertical asymptote at $x=2$, thus $-\infty$ /$+\infty$ left and right limits. Taking the exponential, limits are obviously $0$ and $+\infty$ (think that $e^x$ grows very fast and $e^{-x}=1/e^x$).
The behavior of the function at $2^-$ is very special, as it tends to $0$ extremely fast because of the exponential, and it turns out that the left derivatives of all orders are $0$.
