Re: First Year Uni Calculus Marathon
Sorry about the delay, I was skyping gf. I actually deleted my previous post about bait because I changed my mind. (My initial post was rather rushed and the edit was made shortly after.)
I do believe the limit is the same. The issue of there being a sequence of domain holes accumulating at 0 doesn't affect the usual definition of the limit.
If f is a function defined on domain D (say a subset of R^n) and p is in the closure of D, then lim f(x) =L as x->p does not require that we can find punctured balls about p that map into epsilon balls about L, it is just requires that we can find punctured balls whose intersections with D map into epsilon balls about L.
Apologies for writing this in text. Will be clearer if anything I have said is not understood.
If this wasn't the case, we wouldn't be able to study concepts like continuity on objects like fractals, where no point is interior to the set.
I admit that some courses in single variable calculus might insist that f be defined in a punctured neighbourhood of the point p, but its not really a mathematical issue here and more one of definition.
If one insists that f be defined in punctured neighbourhoods of p then we cannot talk about the limit of f at the endpoints of an interval either, if f is not defined outside of the interval.
Ah, I see. I was (evidently mistakenly) thinking the limit wouldn't exist (or even be meaningful to talk about), because the expression in question (the thing with the t*sin(1/t)'s) isn't even defined in any neighbourhood of 0, is it ('cos it blows up infinitely often around 0 I think).
So speaking of which, regarding your comment in the first of the quotes above, with p being in the closure of D, in this case, is it even possible to make a D for the function (the ((t*sin(1/t))^2 – 1)/(t*sin(1/t)) thing) such that p (the point 0) will be in the closure of D? Because this function isn't well-defined in any neighbourhood of 0, is it?
Basically I think the main thing surprising me is that this function (I think) blows up infinitely often around 0 but can still have a limit (but I don't know much about this so for all I know it's quite common).
Also to show the limit was 2, could you actually just use the change of variable in the limit (x = t*sin(1/t)), or did you need to do a bit more work? Because even if this limit wasn't a counterexample, I do think there are other counterexamples showing we can't just always do a change of variables like this.
Edit: oops, forgot my own Q's function, it was actually ((1+t*sin(1/t))^2 – 1)/(t*sin(1/t)) -_- .
$ existing (for some real number $a$) but not equal to $f^\prime (a)$? (seanieg89 has just shown above that $f^\prime (a)$ exists.) Justify your answer.$)
 we do, however, know that 'direct substitution' of 0 yields the indeterminate form }\frac{0}{0}\text{ again.}\\ \begin{align*}\lim _{t\to 0} \frac{\left(1+t\sin \frac{1}{t}\right)^2 -1}{t\sin \frac{1}{t}}& \stackrel{L'H}{=}\frac{2\left(1+t\sin \frac{1}{t}\right)\cdot \frac{d}{dt} \left(1+t\sin \frac{1}{t}\right)}{\frac{d}{dt}\left(1+t\sin \frac{1}{t}\right)}\end{align*}\\ \text{which is the same thing...}(=2))