Midterm Exam 2 Solutions

# Midterm Exam 2 Solutions - MATH150 Midterm Exam 2 Solutions...

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

MATH150 Midterm Exam 2 Solutions 11/18/05 1. (12 marks) Determine the following limits: (a) lim x →∞ ln(ln x ) ln 2 x (b) lim x 0 sinh( x ) - x x 3 (c) lim x π/ 2 - (cos x ) cos x Solution : (a) lim x →∞ ln(ln x ) ln 2 x = lim x →∞ 1 ln x · 1 x 2 ln x · 1 x = lim x →∞ 1 2 ln 2 x = 0. (b) lim x 0 sinh x - x x 3 0 0 = lim x 0 cosh x - 1 3 x 2 0 0 = lim x 0 sinh x 6 x 0 0 = lim x 0 cosh x 6 = 1 6 . (c) Let’s ﬁrst compute: lim x π/ 2 - cos x ln(cos x ) = lim x π/ 2 - ln(cos x ) sec x = lim x π/ 2 - 1 cos x · ( - sin x ) sec x tan x = lim x π/ 2 - - tan x sec x tan x = lim x π/ 2 - - 1 sec x = lim x π/ 2 - - cos x = 0 Thus lim x π/ 2 - (cos x ) cos x = e lim x π/ 2 - cos x ln(cos x ) = e 0 = 1. 2. (12 marks) What is the maximal area a right triangle can have whose hypotenuse has a length of 2 cm? Carefully justify your answer! Solution : Let the lengths of the sides of the triangle be x , y and 2. The area of the triangle is A = 1 2 xy , where x 2 + y 2 = 4 y = 4 - x 2 . Thus we have to maximize A ( x ) = 1 2 x 4 - x 2 , deﬁned on the interval [0 , 2]. A

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.

## Midterm Exam 2 Solutions - MATH150 Midterm Exam 2 Solutions...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online