ap-calculus-ab None Q20

ap-calculus-ab · USA · -bc_1969-1998_multiple-choice-collection Not Maths Verifying Statements About Probability Properties
20. The statement " $\lim _ { x \rightarrow a } f ( x ) = L$ " means that for each $\varepsilon > 0$, there exists a $\delta > 0$ such that
(A) if $0 < | x - a | < \varepsilon$, then $| f ( x ) - L | < \delta$
(B) if $0 < | f ( x ) - L | < \varepsilon$, then $| x - a | < \delta$
(C) if $| f ( x ) - L | < \delta$, then $0 < | x - a | < \varepsilon$
(D) $\quad 0 < | x - a | < \delta$ and $| f ( x ) - L | < \varepsilon$
(E) if $0 < | x - a | < \delta$, then $| f ( x ) - L | < \varepsilon$
20. The statement " $\lim _ { x \rightarrow a } f ( x ) = L$ " means that for each $\varepsilon > 0$, there exists a $\delta > 0$ such that\\
(A) if $0 < | x - a | < \varepsilon$, then $| f ( x ) - L | < \delta$\\
(B) if $0 < | f ( x ) - L | < \varepsilon$, then $| x - a | < \delta$\\
(C) if $| f ( x ) - L | < \delta$, then $0 < | x - a | < \varepsilon$\\
(D) $\quad 0 < | x - a | < \delta$ and $| f ( x ) - L | < \varepsilon$\\
(E) if $0 < | x - a | < \delta$, then $| f ( x ) - L | < \varepsilon$\\