How does the definition of a limit prove itself if the values (super) close to the value suddenly changes?

Question

So I wondering how the definition of a limit is proved if suddenly when $x \rightarrow a$ the value suddenly changes.

Consider the function and its limit as $x \rightarrow 0$:

$\displaystyle{\lim_{x \to 0}} \frac{\sqrt{x^2+9}-3}{x^2}$

In this example, values "close" to $1$ seem to tend towards $0$, however, "more closer" the values suddenly change and the limit is not defined. So how does the definition of a limit prove itself, how close is close?

Close-up:

Closer-up:

Thanks.

Answer 1

Your "closer-up" graph is simply wrong. The values do not change in the way shown on the graph, and instead continue to smoothly approach $1/6$ like in your "close-up" graph.

The strange behavior of the "closer-up" graph is presumably a result of the program being used to draw the graph not being able to calculate intermediate steps with enough precision in order to get a good approximation to $\frac{\sqrt{x^2+9}-3}{x^2}$ when $x$ is extremely small. In particular, I would guess that small rounding errors in the computation of $\sqrt{x^2+9} - 3$ (which is a number little more than $0$) are getting magnified enormously because they are being multiplied by the very large number $\frac{1}{x^2}$ (i.e. $(\sqrt{x^2+9} - 3) \cdot \frac{1}{x^2}$, which is supposed to be close to $\frac{1}{6}$)

In any case, to rigorously prove that the limit has a certain value, it does not suffice to zoom in on the graph to any level of precision. Instead you must prove that the limit definition is satisfied, so that for all $\epsilon>0$ there exists a $\delta>0$ with the required property. This $\delta$ could be arbitrarily small, possibly too small to show up on whatever graph you draw (i.e., you need to zoom in more to see that the function stays within $\epsilon$ of the limit).

Answer 2

Too long for a comment, but here's how I would evaluate this limit: \begin{align} \lim_{x \to 0} \frac{\sqrt{x^2 + 9} - 3}{x^2} &= \lim_{x \to 0} \frac{(\sqrt{x^2 + 9} - 3)(\sqrt{x^2 + 9} + 3)}{x^2(\sqrt{x^2+9} + 3)} \\ &= \lim_{x \to 0} \frac{x^2 + 9 - 9}{x^2 (\sqrt{x^2 + 9} + 3)} \\ &= \lim_{x \to 0} \frac{1}{\sqrt{x^2 + 9} + 3} \\ &= \frac16. \end{align}

Answer 3

There is nothing strange happening at any point close to $x=0$. The function is perfectly well defined, continuous and approaches the same value from both sides. It looks like this:

Where the point exactly at $x=0$ is undefined. But the limit exists and it's $1/6$ as @littleO shows. The value never "suddenly changes", and the limit is well defined. Here's a nice article that might help: http://www.mathwarehouse.com/calculus/limits/how-to-determine-when-limits-do-not-exist.php