The only was I see a purely algebraic way of finding a limit would work if it ends with

limt_{h -> 0} const. = const.

as in the limit h to 0 of

[math] \frac {(x+h)^2-x^2} {x+h} = \frac {x^2 + 2xh +h^2 -x^2} {x+h} = 2x \frac {x+h} {x+h} =2x [/math]

In that case the solution will be unique, yes.

>>7858022
Could we perhaps work with preimages and open sets?

Dumb weeaboo

>>7858022
>One can get different results via different algebraic manipulations
wut

>One can get different results via different algebraic manipulations.
In a Hausdorff space every limit is unique.

>>7858126
Why are we assuming that we are even in Hausdorff space?

>>7858114
How do I know when I should plug the x in when evaluating a limit, after algebraically manipulating it?

If I plug the x in after barely doing any manipulation, I get one result (usually undefined).
If I plug it in after a lot of manipulation, I get another one.
How do I know which one is correct?

>>7858134
Also, after doing even more manipulation, the result may be entirely different after I plug the x in.

>>7858134
[eqn] \lim_{x \to c} f(x) = f(c) [/eqn]
Is only true is [math] f[/math] is continuous in [math] c[/math].

>>7858060
How did you get from step 2 to step 3?

>>7858133
Why would we make crazy assumptions like that?

>>7858134
>How do I know when I should plug the x in when evaluating a limit,
When you're able to. If you have 0/0 then the form is indeterminate, you need to do more work. Note that an indeterminate form does not necessarily mean the limit does not exist.

>>7858184
Because there are plenty of easy to think about topologies which we can experiment with.

>>7858134
Can you please differentiate between undefined and indeterminate?

>>7858179
I made an error, the denominator must be h, not x+h.
Result is the same.