Question

Determine the limit l for the given a, and prove that it is the limit by showing how to find a d such that |f(x)-l| 11 years ago

Answer 1

\[ \lim_{x\rightarrow 0}{x(3 - cos^2x) = 0} \]

Answer 2

Yes, the limit is zero, but the exercise is to construct a proof of that using delta-epsilon notation.

Answer 3

ugh.... epsilon-delta... imaoutta here!!!

Answer 4

@dpaInc I don't blame you :P

Answer 5

\[ |(x(3 - \cos^xx)) - 0| < \epsilon \]
\[ |(x(3 - \cos^xx)) - 0(3 - \cos^xx))| < \epsilon \]
\[ |(x-0)(3 - \cos^xx)) | < \epsilon \]
\[ |(x-0)) | < \epsilon/|(3 - \cos^xx)| < \delta/3 \]

Answer 6

so, choose delta such that delta < e/3?

Answer 7

I'm still trying to work out how we get this into a proof. Maybe I'm making it harder than it is. The function's continuous, so it's intuitively obvious that there's always a delta for every epsilon. I'm just not good at proof-writing yet. I follow what @experimentX is doing up until the end, I don't see where delta/3 comes from.

Answer 8

shouldn't it be delta < epsilon/2 ?

Answer 9

but that's just me.. i get things wrong all the time!!! :)

Answer 10

I think I might get where the delta/3 is coming from, but I should point out, it's not cos^2(x), it's cos(x^2). Not positive if that's relevant or not.

Answer 11

limit implies, |f(x) - L| < e implies the existence of |x-a| < d, both e and d are arbitrarily chosen, and it says that when one decreases the other also should,

cos x^2 doesn't matter either.

Answer 12

I didn't think it did at first, which is why I only just mentioned it, but if it doesn't then I'm still lost as to where the delta/3 is coming from.

Answer 13

yea shouldn't matter. but shouldn't delta be the bigger one of (e/2 or e/3) because if cos(x^2) = 1, 3-1=2

Answer 14

Alright, to the back of the book we go.

L=0. For all x we have \[|cos(x^2)|\le1\] so \[|3-cos(x^2)|\le4\] and thus \[|f(x)-0| = |x| \cdot |3 - cos(x^2)|\le4\cdot |x|\]
So we can take \[\delta =\epsilon /4\]

I... kind of get it.