Limits in math? I really need this explained easy, please.

[Moosewala]

Hello guys,,
Limits are: "What do I get if x is reeeaally close to (some number or infinity)"

Then you notice a pattern. For x approaching infinity in 1/x your answer gets smaller the bigger x is. Hmm then if x was an insanely big number then my answer would be insanely small. Which is pretty close to 0.

[HallsofIvy]

What you wrote is not very clear and I don't see a question!
Yes, you can say that "[tex]\lim_{x\to a} f(x)= L[/tex]" is very roughly saying "If x is really near a then f(x) is really near L". You can say more- The closer x gets to a, the closer f(x) gets to L.

Notice that this does NOT say that f(a)= L! (I almost wrote that it was, above, without thinking.) For example, if f(x) is defined to be "3x- 5 for all x except x= 2" and f(2)= 10, then [tex]\lim_{x\to 2} f(x)= 6- 5= 1[/tex] even though f(2)= 10. Of course, if it happens that [tex]\lim_{x\to a} f(x)= f(a)[/tex] that's a very nice property and we like to work with such functions, which we say are "continuous".

You appear to be talking about the specific problem of [tex]\lim_{x\to\infty} \frac{1}{x}= 0[/tex]. "Near" infinity is a little different from ""near" a for finite a. For finite a, we would say that "x is near a" if [tex]|x- a|< \delta[/tex] where [tex]\delta[/tex] is an arbitrarily small positive number. For x going to infinity we say, instead, that x> A where A is an arbitrarily large positive number.

In particular, to prove that [tex]\lim_{x\to\infty} \frac{1}{x}= 0[/tex], we have to say that "given [tex]\epsilon> 0[/tex] there exist R> 0 such that if x> R then [tex]\frac{1}{x}< \epsilon[/tex]. We do that by noting that if [tex]\frac{1}{x}< \epsilon[/tex] then, since x> 0, [tex]1< x\epsilon[/tex] so [tex]\frac{1}{\epsilon}< x[/tex]. Take [tex]R= \frac{1}{\epsilon}[/tex] and reverse the argument above.