I haven't ever taken an analysis course, so maybe that's where I would really learn this, but I've always wondered why it's okay to do this when evaluating a limit. I guess it's the case that there is a theorem which says that the limit of a rational function as $x\rightarrow a$ is equal to the l...
I understand how it is legit, from this coursera calculus course I'm taking, the lecturer says, both functions are not same but similar and when cancelling common factors you get a function defined where the original function was undefined.
However, it seems like black manage to me how removing common factors from a function results in a similar function without a hole ? $frac{(x-1)(x+1)}{x-1}$ = $frac{(x+1)}$
Is this always the case, when cancelling common factors of a function I get a function that plots a similar graph?
@gideon whenever you cancel (x-a) from numerator and denominator of a rational function, you remove the removable singularity x=a from the graph (i.e. you fill in the "missing point"), and otherwise it is unchanged
@anon Offshoot question : the article says "Taking a power series expansion" looks a little like taylor series, are they the same thing power series/taylor series ?
usually. technically power series can be more general (for example, formal power series need not converge anywhere or define any function (they are used to encode combinatorial information, for example), you can have p-adic expansions etc.) and if you include negative powers you get Laurent series, not Taylor series
If removable singularities did not exist, then will nullify the notion of the limit? I mean, then a limit of a function with just be what the function is at that point. ?
ok, so forgive the ignorance I'm trying to learn calculus on my own. I'll tell you what the confusion is in my head, somehow I see the concept of a limit only standing for cases where there is this removable singularity, where you have to do something special. Otherwise the limit at x->a is f(a)
I'm seeing the limit as being something special only when there is a hole in the function and then yes it makes sense that although the function doesn't have a value there it's limit can exist.
but only when there is a hole. So does the whole limit thing harbour on the fact that there are holes in functions sometimes.
you can also take one-sided limits, or limits at infinity. these are ubiquitously useful for asymptotic analysis. also, limits can be nontrivial for a function at a place where it has a hole that has been "filled-in incorrectly" (for example consider f(x)=1 when |x|>0 and f(x)=0 when x=0).
you also need limits to define derivatives and integrals in the basic calculus sense
so think about very small numbers. x=0.001, x=0.000001, x=0.000000001, etc.
for all those, f(x) = 1, because |x| > 0
but at x = 0, we have f(x) = 0.
and that works for both sides.... x=-0.001, x=-0.000001, x=-0.000000001, etc.
you'd think that f(0) should be 1, because you can get really close on either side and it's always 1, as close as you want. but we defined the function so that f(0) = 0.
the basic idea behind continuity is that behavior like that is kind of crappy, you don't want to be dealing with functions that just jump around unpredictably all of a sudden. if a function doesn't do that, we call it continuous, i.e. the value of the function is equal to its limit at every point. it's one way of saying it's a "nice" function.
I was trying to do $G/C(G)$ cyclic $\implies G$ abelian in my head and I was so stuck. And then when I finally pulled out a pad of paper it was so easy.
The whole time while walking and thinking I was like "Well, let's say I have an inner automorphism $\psi_h$ that generates the group of inner automorphisms..."
And then I felt so dumb because there's no need to be so fancy.
Fine, if $G$ has a subgroup $H \neq G$ of finite index, then it has a normal subgroup $\neq G$ of finite index.
Simple groups are a good point, though, since it shows we can't do certain various trivial ideas, like "intersection of all normal subgroups containing $H$" (since in a simple group this will just be $G$).
