# Disproving a limit from the formal definition

• Mar 18th 2011, 09:37 AM
moses
Disproving a limit from the formal definition
Hello :)

I'm having a lot of trouble with this proof:

Prove that the limit of f(x) = (x+1)/3 as x approaches 1 is NOT 2/3 + 10^-100

I've tried using the definition of the limit with epsilon as 10^-100/2 to get to a contradiction, but somehow it's not working out...any help would be greatly appreciated.
• Mar 18th 2011, 10:10 AM
NOX Andrew
Are you allowed to use the formal definition of a limit to prove $\displaystyle \lim_{x \to 1} \dfrac{x + 1}{3} = \dfrac{2}{3}$, and in doing so disprove $\displaystyle \lim_{x \to 1} \dfrac{x + 1}{3} = \dfrac{2}{3} + 10^{-100}$?
• Mar 18th 2011, 01:04 PM
moses
No, I'm not allowed to assume that the limit is unique.
• Mar 18th 2011, 02:51 PM
tonio
Quote:

Originally Posted by moses
Hello :)

I'm having a lot of trouble with this proof:

Prove that the limit of f(x) = (x+1)/3 as x approaches 1 is NOT 2/3 + 10^-100

I've tried using the definition of the limit with epsilon as 10^-100/2 to get to a contradiction, but somehow it's not working out...any help would be greatly appreciated.

$\displaystyle{\left|\frac{x+1}{3}-\frac{2}{3}-\frac{1}{10^{100}}\right|<\epsilon\Longrightarrow \left|10^{100}(x+1)-2\cdot 10^{100}-3\right|<3\cdot 10^{100}\epsilon$ .

Since the LHS in the last inequality is negative for values of x very close to 1, we get

that this inequality is:

$\displaystyle{3+10^{100}-10^{100}x<3\cdot 10^{100}\epsilon\Longrightarrow x>\frac{3+10^{100}-3\cdot 10^{100}\epsilon}{10^{100}}$ .

If we now choose $\displaystyle{\epsilon=\frac{1}{3\cdot 10^{100}}$ , the last inequality becomes:

$\displaystyle{x>\frac{3+10^{100}-1}{10^{100}}=1+\frac{2}{10^{100}}$ .

The above already shows that there can't be any

$\displaystyle{\delta>0\,\,s.t.\,\,|x-1|<\delta\Longrightarrow \left|\frac{x+1}{3}-\frac{2}{3}-\frac{1}{10^{100}}\right|<\frac{1}{3\cdot 10^{100}}$ .

Tonio
• Mar 18th 2011, 03:32 PM
moses
Thanks a lot, but I don't quite grasp the last step...how does the bound on x lead to the conclusion?
• Mar 18th 2011, 04:16 PM
tonio
Quote:

Originally Posted by moses
Thanks a lot, but I don't quite grasp the last step...how does the bound on x lead to the conclusion?

Which "bound x"? As we saw, x has to be greater than 1.0000.........02 (100 zeroes) in order to fulfil the requirement of

the limit's definition. But for any positive delta, there will be values of x s.t. |x - 1|

is less than the above

value and, thus, the abs. value...etc won't be less than that epsilon...

Tonio
• Mar 18th 2011, 04:48 PM
moses
Okay, I think I see...in order for the requirement to be fulfilled, x must be greater than 1.0000...02, but no matter what positive delta is given, there will always be values of x that both fulfill |x-1| < delta and x < 1.0000....002, in other words there will always be values at which the function is more than a distance of epsilon away from 2/3+10^-100.
• Mar 18th 2011, 06:44 PM
tonio
Quote:

Originally Posted by moses
Okay, I think I see...in order for the requirement to be fulfilled, x must be greater than 1.0000...02, but no matter what positive delta is given, there will always be values of x that both fulfill |x-1| < delta and x < 1.0000....002, in other words there will always be values at which the function is more than a distance of epsilon away from 2/3+10^-100.

Yup, you put it clearer than I did.

Tonio