Confusion

[bezgin]

How can we prove that the lim (x->1) x^2 + 2 is NOT equal to 2.999? (example I made up right now) At the end of each proof we find a relation between epsilon and delta. What does it mean?

[MiGUi]

You cant take back the 2 as a constant, and the limit of x² will be 1

[Galileo]

Since
\lim_{x\rightarrow 1}x^2+2=2.999 \iff \lim_{x\rightarrow 1}x^2=0.999
we can work with the righthanded expression. (it's easier).
We'll have to find an \epsilon>0 such that for any \delta>0 we have |x-1|<\delta AND |x^2-0.999|>\epsilon

EDIT: Removed 'proof', because of errors.

[jatin9_99]

my friend it's real simple; as the 2 won't be effected by the limit, so only part affected by limit is x^2 when x->1

although u are write it might go to 2.999 when it approach from left, but u need to consider another fact when limit approach from right than it will be 3.00001 that's how we prove that because u don't specify as limit apporaches from right or left

[spacetime]

Proving that \lim_{x\rightarrow a} f(x) = b means that it should be possible, by choosing a \delta , to make the make the difference in values of f(x) and b smaller than any positive number \epsilon , for all the values of |x-a| less than \delta .

For 3, you can show this to be true. But for 2.999, although you can choose a number for which the difference is smaller than any positive number, but this will not be true for all the |x-a| smaller than that chosen number.

That is the reason why 2.999 isn't the limit of the function.


spacetime
www.geocities.com/physics_all/index.html

[bezgin]

These replies make think, once again, what does it mean to be "close"?
Can't we find a relation between epsilon and delta by doing the operations abs(x^2 + 2 - 2.999) < epsilon and abs(x-1) < delta?
I'm really stuck with this issue for two weeks and think I will not understand the concept of the formal definition of a limit. I've tried nearly anything on the web :(

[arildno]

Sure, here's how:
|x^{2}+2-2.9991|=|(x-1)(x+1)+r|,r=0.0009
1) Assume x>1:
Then r+(x-1)(x+1)>r>0, so:
|(x-1)(x+1)+r|>r.
Choose in this case, \epsilon<r
2)Assume x<1.
Then, you can find \delta, so that (1-x)(x+1)<\frac{r}{2}
Hence, by the triangle inequality, we have:
|x^{2}+2-2.9991|=|(x-1)(x+1)+r|\geq(r-(1-x)(x+1))\geq(r-\frac{r}{2})=\frac{r}{2}

Hence, the choice \epsilon<\frac{r}{2} is always out of reach, so 2.9991 cannot be the limit.