{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MIT18_014F10_pr_ex3_sols

# MIT18_014F10_pr_ex3_sols - Practice Exam 3 Solutions...

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

0 Practice Exam 3 Solutions t 3 + t 1+ t 2 dt Problem 1. Evaluate Solution The problem can be simplified as t 3 + t = t ( t 2 + 1). Then by substitution with u = t 2 + 1 and thus du = 2 t dt we have t t 2 + 1 dt = 1 u 1 / 2 du = 1 u 3 / 2 + C = 1 ( t 2 + 1) 3 / 2 + C. 2 3 3 2 Problem 2. Evaluate 5 x 3 x 9 dx 3 Solution We begin by making the substitution x 2 9 = u . Then 2 xdx = du and x 2 = u + 9. Substituting in, we get u 3 / 2 16 16 16 du 1 1 2 18 1 ( u +9) u 3 / 2 +9 u 1 / 2 du = 5 / 2 16 5 / 2 +3 16 3 / 2 + = u u = . · 2 2 2 5 3 5 0 0 Problem 3: Suppose that lim x a + g ( x ) = B = 0 where B is finite and lim x a + h ( x ) = 0, but h ( x ) = 0 in a neighborhood of a . Prove that g ( x ) h ( x ) lim x a + = . Solution Let M R + and set = 1 / (2 M ) > 0. By hypothesis, there exist δ 1 , δ 2 such that | g ( x ) B | < | B | / 2 if 0 < x a < δ 1 and | h ( x ) | < | B | if 0 < x a < δ 2 . Choose δ = min { δ 1 , δ 2 } . Then for 0 < x a < δ , | g ( x ) | > | B | / 2 and | h ( x ) | 1 > 1 / ( B � ). Thus | | | g ( x ) | > | B | = 1 = M.

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 ]}