English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

prove that Lt x->inf. x{sqrt(x^2+sqrt(x^4+1))-x sqrt(2)}=0
and also prove that Lt x->inf.{cos(sqrt(x+1))-cos(sqrt(x))}=0

2007-11-06 05:17:01 · 1 answers · asked by Anonymous in Science & Mathematics Mathematics

1 answers

For the first limit, multiply and divide the term inside the curly brackets by sqrt(x^2+sqrt(x^4+1))+x sqrt(2). If you simplify the numerator you should get
sqrt(x^4+1) - x^2. I would also deal with the factor of x outside the curly brackets at this point: factor an x from the denominator, which will give you

x[sqrt(1 + sqrt(1 + 1/x^4)) + sqrt(2)]

which you can use to cancel the x outside the curly brackets.

To handle the sqrt(x^4+1) - x^2 numerator, use the same sort of trick as before: multiply numerator and denominator by sqrt(x^4+1) + x^2; after simplifying you'll end up with 1 in the numerator. Now you should be able to show that the limit is zero.

This "trick" of using
sqrt(A) - sqrt(B) = [sqrt(A) - sqrt(B)][sqrt(A) + sqrt(B)]/[sqrt(A) + sqrt(B)]
= (A-B)/[sqrt(A) + sqrt(B)]

is often helpful in evaluating limits involving differences of square roots.

For the second limit, use the trig identity

cos(C) - cos(D) = 2 sin((C+D)/2) sin((D-C)/2)

You'll also need that trick employed twice in the first limit.

2007-11-06 06:01:39 · answer #1 · answered by Ron W 7 · 0 0

fedest.com, questions and answers