{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# ans4 - MAA 4200 Homework Solutions Sequential Limit Theorem...

This preview shows pages 1–2. Sign up to view the full content.

MAA 4200: Homework Solutions June 19, 2009 Sequential Limit Theorem = Theorem 2 . 1 Algebra of Limits Theorem = Theorem 2 . 4 [2.12] Prove that if lim x x 0 f ( x ) = L then lim x x 0 | f ( x ) | = | L | . Preliminary Work: | | f ( x ) | - | L | | ≤ | f ( x ) - L | for all x D by the triangle inequality. Proof: Let > 0. By definition of limit, there exists δ 1 > 0 such that | f ( x ) - L | < for 0 < | x - x 0 | < δ 1 . Choose δ = δ 1 and let 0 < | x - x 0 | < δ . Then | | f ( x ) | - | L | | ≤ | f ( x ) - L | < . [2.13] In Example 2.6 on p. 71 it is proved that lim x x 0 [ x ] exists iff x 0 is not an integer. Thus, since lim x x 0 exists for all x 0 R , the limit lim x x 0 ( x - [ x ]) exists for all x 0 / Z by the Algebra of Limits Theorem. We can also apply the Algebra of Limits Theorem to show that lim x x 0 ( x - [ x 0 ]) does not exist for x 0 Z . Let f ( x ) = x - [ x ]. Proof by contradiction: suppose lim x x 0 f ( x ) exists for some x 0 Z . Then lim x x 0 [ x ] = lim x x 0 ( x - f ( x )) exists by the Algebra of Limits Theorem, which is a contradiction. [2.14] Show that lim x 2 f ( x ) = 16 and the limit does not exist at any other point.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}