Proof:Limit of x at 0 is 0

From CS2800 wiki
[math]\href{/cs2800/wiki/index.php/%5Clim}{\lim}_{x → 0} f = 0 [/math] where [math]f(x) := x [/math].
Proof:
We want to show that [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) - 0| \lt ε [/math].


Choose an arbitrary [math]ε \gt 0 [/math]. Let [math]δ \href{/cs2800/wiki/index.php/Definition}{:=} ε [/math] ([math]δ \gt 0 [/math] since [math]ε \gt 0 [/math]). Choose an arbitrary [math]x \href{/cs2800/wiki/index.php/%E2%88%88}{∈} \href{/cs2800/wiki/index.php?title=%E2%84%9D&action=edit&redlink=1}{ℝ} [/math], and assume that [math]0 \lt |x - x_0| \lt δ [/math]. We want to show that [math]|f(x) - 0| \lt ε [/math].


Well, by definition, [math]f(x) = x [/math], and by assumption, we have [math]0 \lt |x - x_0| \lt δ [/math]. But [math]ε = δ [/math] by definition; therefore [math]|f(x) - x_0| = |x - x_0| \lt δ = ε [/math] as required.