Why is $0/\lim_{x\to0} x$ undefined, when $\lim_{x\to2} (x^2-4)/(x-2)$ is defined?

Question

For $\lim\limits_{x\to2}\ (x^2-4)/(x-2)$, we are able to cancel out $x-2$ and rewrite it as $\lim\limits_{x\to2} x+2 $ But in maths, we are not able to cancel out $0$ values so $x-2$ is not zero, and it totally makes sense. $x$ is aproaching to 2 but it is NEVER 2, it's extremely close to 2 but never 2.

Here is my main problem, $0/\lim_{x\to0}x$

I am told that this is undefined but I don't understand why. I mean, as we have done above, $x$ is something extremely close to $0$ but its never $0$ and '$0$ divided by something not $0$ ' is not undefined. It is $0$, so why isn't the answer $0$ then?

Answer 1

$ \lim_{x\to0} 0/x $ is equal to zero for the reasons you suggest.

note $ \lim_{x\to0} x =0$

so $0/ \lim_{x\to\infty} x = 0/0$ is undefined

Answer 2

This is the mathematical definition of limit- we say $ \lim_{x \to c } f (x) = L$ if for each $\epsilon > 0$ there exist a $\delta > 0$ such that $0< |x-c|< \delta \implies |f(x) - L | < \epsilon$. By this definition we can easily avoid $\frac 00 $ confusion.

The above definition is only abstraction of our intution. Suppose a function has a limit at a point it can be made arbitrarily close to a number which is called limit value of the function at the given point . In your case suppose you want make distance between $f(x) = \frac{x^2-2}{x-2}$ and $4$ less than $\frac{1}{1000}$ then if you consider $x \in (2 - 10^{-3} , 2 + 10^{-3})$ then you observe that $|f(x) - 4 | = |x-2| < \frac{1}{1000}$. How wonderful it is! Similarly you can take any number ($\epsilon$ ) and find an interval for values of $x$ ($|x-2| < \delta$) such that distance between $f(x)$ and $4$ is less than $ \epsilon$.

Thus we have avoided confusion about $0/0$ form in limits. If you consider $ \frac{0}{lim_{x \to 0 } x }$ this is little different than the above. We have $lim_{x \to 0 } x = 0$ and obviously $0/0$ is undefined (why? see here).