Determined or not? the function $\dfrac {2x}{3x-\sqrt{x} }$ is not derterined for values of $x$ equale or samller than zero, though when I take the limit $ \lim_{x \to 0^+} \dfrac {2x}{3x-\sqrt{x} }$ the output is zero as though the function gets the value $(0,0)$ at that point, even though we know that's not the case....how come? 
 A: It is actually the whole point of defining limits. The reason is to study the behaviour of functions in messy and difficult positions. If you know the definition of a limit you will note that the actual value of the function at the point in which the limit is evaluated is redundant. Say we require the value of the limit of a function at the point $a$. The function need not be defined at $a$ necessarily. But if the limit $L$ exsists then what we need to make sure is the fact that $f(x)$ is arbitrarily close to $L$ when $x$ is arbitrarily close to $a$. For an exmple the function $ f(x) = \dfrac{x^2 - x}{x - 1} $ is nothing but the function $f(x) = x$ except for the point $x = 1$ where it is undefined. You will note that by the $\epsilon$-$\delta$ definition of a limit we can prove the function is arbitarily close to $1$ when $x$ is arbitrarily close to $1$. That is all we need to assert that $\lim_{x \to 1}  \dfrac{x^2 - x}{x - 1} = 1$
A: Hint: Note that while $x\to 0^+$, $x$ is not standing on $0$ on the real line. So we have $$x=\sqrt{x}\sqrt{x}, ~x>0$$ and are allowed to omit the irritating parts from numerator and denominator. 
A: Limit can be defined at $x_0$ regardless of the value at $f(x_0)$. Indeed, the function has a limit, but it's not continuous at $x_0$.
Also, consider this: Classification of discontinuities
A: perhaps refresher on "removable discontinuities" would be helpful. http://www.drensplace.com/Math/Calculus/page0/page20/page37/page37.html shows 2 types of removable discontinuities, one with $f(a)$, on the other with $f(a)$ not defined.
A when calculating a limit at $a$, we are unconcerned about how it is defined at $a$. What is important is how the function behaves NEAR $a$ (sorry, was that too loud?).
A limit is really just a game. Suppose we suspect that $\displaystyle\lim_{x\to a}f(x)=L$. You get to pick a number $\epsilon\gt 0$. If I can pick a number $\delta\gt 0$ such that $L-\epsilon\lt f(x)\lt L+\epsilon$ whenever $a-\delta\lt x\lt a+\delta$ (except possibly at $x=a$, because $f(a)$ may be undefined), then I win a jellybean. If I can always win a jellybean, no matter what value you pick for $\epsilon$, then we say that $L$ is the limit.
