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Why don't these limits exist?

$$\lim_{x\to 0}\frac1{x^2}$$

Aren't we here just approaching $x$? And, thus will not get $0$ and avoid the division on "0"?

And, what about this limit?

$$\lim_{x\to 0}\frac{|x|}{x}$$

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We are approaching x, but the value of 1/x tends to plus or minus infinity. (depends if you approach x=0 from below or above). 1/.001=1000, which is big. – Thomas Rot Mar 21 '11 at 8:50

It's not clear to me what you intended the first limit expression to be. In the case of the second one, consider the graph of $y=\frac{|x|}{x}$:

plot of Abs[x]/x

As $x\to 0^+$ (that is, as $x$ goes to $0$ from the positive/right side), $\frac{|x|}{x}\to 1$, but as $x\to 0^-$ (that is, as $x$ goes to $0$ from the negative/left side), $\frac{|x|}{x}\to -1$. Since these two "one-sided" limits do not agree, we say that $$\lim_{x\to 0}\frac{|x|}{x}$$ does not exists.

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Thanks for your reply. So, is that since approaching 0 once from left and once from right does not lead to the same value, we say it doesn't exist? – Simplicity Mar 22 '11 at 8:18
@SWEngineer: More or less, yes, if the two one-sided limits are not equal ($x\to a^+$, $x\to a^-$), the limit ($x\to a$) does not exist. This is not, however, the only way for a limit to not exist. A typical definition is along the lines of $\underset{x\to a}{\lim} f(x)$ exists if and only if the value of $f(x)$ gets arbitrarily close to some number $L$ (the limit) as $x$ gets arbitrarily close to $a$. This means that, as Arturo said in his answer, $\underset{x\to 0}{\lim}\frac{1}{x^2}$ does not exist because $\frac{1}{x^2}\to\infty$ as $x\to 0$ and $\infty$ is not a number. – Isaac Mar 22 '11 at 8:27

In these instances, the phrasing 'does not exist' really has two related meanings, one for each of your examples.

  • for the first example, whether you go from the left or the right, you approach a very large number (what is usually called infinity' or '$\infty$'). And this 'number' (it's a bit questionable to call it a 'number') is not manipulable by our usual means (it follows some rules, bu not all. So, to avoid allowing use of it afterwards, we say 'it does not exist'

  • for the second example, the limit value is different when going from opposites directions. Since there's no good way to choose between them, and we really don't think of passing along pairs of numbers, we just say 'the limit does not exist'.

So in some sense these answers, $\infty$ and the pair {$\infty$ and $-\infty$), exist, but they are not really usable by our usual algebraic manipulations, so we preclude worrying about them by saying 'they don't exist'.

I have a tendency to say that the first example is $\infty$ (because there are instances where you can manipulate that as a value) and the second is 'undefined' (because it is a bit harder to manipulate a function that sometimes returns two values, when really you want only one).

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The problem with the first one is not division by zero. The problem with the first one is that as $x$ approaches $0$, the value $\frac{1}{x^2}$ gets larger in size; it can get larger than anything you care to specify ahead of time, by simply taking $x$ small enough. So although we never have to actually divide by $0$, the problem is that the values are not approaching anything: they are "going to $\infty$" (meaning, growing without limit).

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$f(x)\to\infty$ as $x\to0$ if we can find some $k\gt0$ such for any $A\gt0$, $f(x) \gt A$ when $|x|\lt k$ from this the answer for the fist one is $f(x)\to\infty$ (the limit exists in this particular definition)..see Differential Calculus by Ferrar

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