Why doesn't a limit exist when both the numerator and denominator don't approach 0? For example, $\lim\limits_{x\to5}\frac{x^2-5x+6}{x-5}$ does not exist because $\lim\limits_{x \to 5} x^2-5x+6=6$ and $\lim\limits_{x \to 5} x-5=0$.
 A: The limit in that case is $\infty$ from the right, and $-\infty$ from the left. The intuition is that as you divide a "fixed" positive number by smaller and smaller quantities, the result grows large unboundedly.
In the other hand, if both the numerator and denominator approach $0$ the same argument doesn't work. In fact, you will find examples where both the numerator and denominator approach $0$ and the limit can either exist and be equal to any number (including $0$), or the limit can equal $\pm\infty$, or it may not exist. That's why we say that $0/0$ is an indeterminate.
A: A simpler example than yours is $\lim\limits_{x\to5}\frac{6}{x-5}$
and one even simpler is $\lim\limits_{x\to5}\frac{1}{x-5}$. One still even simpler is $\lim\limits_{x\to0}\frac{1}{x}$. 
Whatever $\delta>0$ you may choose, $\exists K>0$ such that $\forall x $ with $0<\lvert x\rvert<\delta$, it results $\left\lvert\frac{1}{x}\right\rvert>K$. Indeed it is sufficient to choose $K>\frac{1}{\delta}$
A: When one takes the limit as $x\to 5+$ the ratio tends to $\infty$ whereas taking limit as $x \to 5-$ the ratio goes to $-\infty$ here in this particular example. Therefore the limit does not exists for this example.
A: Overarching statement on limits:
When both the numerator $f(x)$ and denominator $g(x)$ of a rational function $r(x) = f(x)/g(x)$ approach $0$ as $x$ approaches some point $a$, the limit of $r$ is said to be of the indeterminate form of type $0/0$.
Here however, your limit is $+\infty$, the statemetn above however, is used for those limits that are indeterminate, which I think you were alluding to in your opening sentence.
