Proof:Limit of x at 0 is not 1

Claim (see proof):

[math]\href{/cs2800/wiki/index.php/%5Clim}{\lim}_{x → 0} f \neq 1 [/math] where [math]f(x) := x [/math].

Proof:

The logical negation of

[math]\href{/cs2800/wiki/index.php/%E2%88%80}{∀} ε \gt 0, \href{/cs2800/wiki/index.php/%E2%88%83}{∃} δ \gt 0, \href{/cs2800/wiki/index.php/%E2%88%80}{∀} x, \text{ if } 0 \lt |x - 0| \lt δ \text{ then } |f(x) - 1| \lt ε [/math]

is

[math]\href{/cs2800/wiki/index.php/%E2%88%83}{∃} ε \gt 0, \href{/cs2800/wiki/index.php/%E2%88%80}{∀} δ \gt 0, \href{/cs2800/wiki/index.php/%E2%88%83}{∃} x, 0 \lt |x - 0| \lt δ \text{ and } |f(x) - 1| \not\lt ε [/math]

Let [math]ε = 1/2 [/math]. Choose an arbitrary [math]δ \gt 0 [/math].

Let [math]x [/math] be the minimum of [math]δ/2 [/math] and [math]1/2 [/math]. Note that [math]0 \lt x [/math], [math]x \leq δ/2 [/math] and [math]x \leq 1/2 [/math] (you can prove this using case analysis).

Then we have [math]0 \lt x \leq δ/2 \lt δ [/math]. Moreover, [math]x ≤ 1/2 [/math] so [math]f(x) - 1 \leq 1/2 - 1 \leq -1/2 [/math], and therefore [math]|f(x) - 1| \geq 1/2 = ε [/math], which is what we were trying to prove.