{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Assignment3Solution(1804)

# Assignment3Solution(1804) - MATH1804/Solution to...

This preview shows pages 1–3. Sign up to view the full content.

MATH1804/Solution to AS3/ZW/2009-10 THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1804: University Mathematics A Solution to assignment 3 08/10/2008 1. Solution : (a) lim x ! 1 5 x 4 ° 4 x 2 ° 1 10 ° x ° 9 x 3 = lim x ! 1 20 x 3 ° 8 x ° 1 ° 27 x 2 = ° 3 7 ; (b) lim x ! 1 2 x 2 ° (3 x + 1) p x + 2 x ° 1 = lim x ! 1 4 x ° 9 2 x 1 = 2 ° 1 2 x ° 1 = 2 1 = ° 1; (c) lim x !1 x tan 1 x = lim x !1 tan (1 =x ) 1 =x = lim x !1 cos ° 2 (1 =x ) ± ° ° x ° 2 ± ° x ° 2 = lim x !1 cos ° 2 (1 =x ) = 1; (d) lim x ! 0 sin 9 x tan 27 x = lim x ! 0 9 cos 9 x 27 cos ° 2 27 x = lim x ! 0 1 3 cos 9 x cos 2 27 x = 1 3 ; (e) lim x ! 0 sin x cos x x ° x 2 = lim x ! 0 cos 2 x ° sin 2 x 1 ° 2 x = 1 . 2. Solution: Let f ( z ) = sin z for any z 2 R . For any x , y 2 R , since f is continuous and di/erentiable on the whole real line, by Mean Value Theorem, there exists a number c lying between x and y , such that f ( x ) ° f ( y ) = f 0 ( c ) ( x ° y ) , or equivalently, sin x ° sin y = (cos c ) ( x ° y ) . Therefore we have j sin x ° sin y j = j cos c j j x ° y j . Since ° 1 ² cos c ² 1 , j cos c j ² 1 , which implies that j sin x ° sin y j ² j x ° y j . 1

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

View Full Document
3. Let x to be the dimension of the equal squares cut from the rectangular sheet and let f ( x ) to denote the volume of the box.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

Assignment3Solution(1804) - MATH1804/Solution to...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online