0

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?

Ottavio
  • 2,287
Dogukan
  • 11
  • 3
    $\lim_{x \to 0}x$ is definetively equal to $0$, so, you cannot divide by $\lim_{x \to 0}x$. – azif00 Dec 25 '20 at 20:55

2 Answers2

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

shai horowitz
  • 2,086
  • in that sense, limx→0 (x2−4)/(x−2) is 0 too – Dogukan Dec 25 '20 at 21:27
  • @Dogukan the problem is your taking the limit of the wrong thing, you want the limit on the outside rather then in the denominator. if its in the denominator then the limit is equal to zero so the whole expression =0/0 which is undefined. – shai horowitz Dec 26 '20 at 00:36
  • Thanks for answering, what makes the difference tho, why one is 0 and one is undefined ? – Dogukan Dec 26 '20 at 08:12
  • 1
    The idea of 'x is close to zero but never zero' only applies to expressions inside the limit, and it is only for the purpose of calculating the limit. Once you've calculated the limit, it's just a number... x has nothing to do with it anymore. For example $2+\lim_{x\to0}x$ is exactly equal to $2+0=2$. It's not 'very close to' 2. It exactly equals 2. – Ameet Sharma Dec 26 '20 at 10:17
  • @Dogukan its an order of operations thing. $0/(\lim_{x\to0}x)$ – shai horowitz Dec 26 '20 at 20:06
0

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).

Infinity_hunter
  • 5,369
  • 1
  • 10
  • 30