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I'm not 100% sure what the " $\epsilon - \delta$ " method is, but it's trivial to calculate the limit as x approaches zero from the positive side. At some very, very small positive number, $\epsilon$ , the function value is 1. No matter how small $\epsilon$ becomes, as long as it's positive, the function value is always 1. Hence the limit from the positive side is 1.

Meanwhile, it's also trivial to calculate the limit from the negative side. Using the same arguments as above, but with some small number $-\epsilon$ , it's easy to see that the limit from the negative side is -1.

Since the limit is different, depending whether you approach zero from the positive side or from the negative side, the actual limit is undefined.

For a function to have a limit you must be able to find a range of x values that give a value for f(x) that is within some arbitrarily small number of the limit L. For example, consider the function y = x^2. It's limit at x = 2 is 4, because I can give you a range of x values (not just a single value) that gets within whatever accuracy you want of 4. If you say: what values of x get me f(x) within 0.01 of 4, I can reply that this is satisfied for x = 2 plus or minus 0.0025. Notice that the "plus or minus" value of x must be something other than plus or minus 0.

But for the function you provided you can't identify a value for the limit that satisfies that. Suppose you say the limit is equal to zero. So I ask what range of x values gets within 0.01 of 0, and your response would have to be that there is no range of x values, there's only one value that satisfies that, namley x =0. Hence this function doesn't satisfy the requirement that there be a range of such values that meets the criteria.

In more formal terms: the definition of the limit says that for any $\epsilon > 0$ there must be a value $\delta > 0$ such that $f(x\pm\delta)$ falls within $L \pm \epsilon$ . For the function you provided there is no $\pm \delta$ for $\delta > 0$ that works. Hence the limit doesn't exist at x = 0.

Quote:

Originally Posted by jcaron2

Since the limit is different, depending whether you approach zero from the positive side or from the negative side, the actual limit is undefined.

True enough. But this may not be a sufficient condition to show the limit doesn't exist. You can have a function where the limit approaching 0 from the negative side is the same as the limit approaching from the positive side, and yet the limit at 0 doesn't exist. Consider:

f(x) = 0 for x<0
f(x) = 1 for x = 0
f(x) = 0 for x>0

The epsilon/delta technique shows that the limit at x=0 does not exist, even though the limits as calculated fom either side are the same.