(March 7, 2018 at 8:53 am)robvalue Wrote:(March 7, 2018 at 8:42 am)polymath257 Wrote: Technically, here, the limit doesn't exist. If the limit *does* exist, the only possibility is what you gave: a removable discontinuity.
And there are many ways a limit can fail to exist:
1. A jump discontinuity (as above)
2. A vertical asymptote (where the limit is infinite). F9x)=1/x does this as x->0.
3. Even the one-sided limits can fail to exist through oscillation. f(x)=sin(1/x) does this as x->0.
Yes, you are correct, the overall limit at X=1 does not exist because the two sides don't match.
Your example 3 is very interesting, I remember it from my lectures a long time ago. They would call it "not well behaved", informally.
That example is used for a large number of other strange examples in upper level courses. So, for example,
f(x)=x*sin(1/x) for x!=0 and f(0)=0 defines a *continuous* function. The limit as x->0 is 0, which is f(0). In essence, the x in front of the sin forces things to go to 0.
Should we start talking about derivatives?
