how do you solve the limit of (x+sinx) / (x+cosx) if x is approaching positive infinity?

Answer 1

since both sin x & cos x have max values of 1, as x--->∞ (x+sin x) & (x+cos x) both aproach x so
(x+sin x)/(x+cos x)-->x/x=1

Answer 2

=(1+sinx/x)/(1+cosx/x)
As sin and cos are bounded sinx/x and cosx/x have limit 0
so the limit is 1

Answer 3

lim x-->0 2+ 3 sin x /(x^3+a million) replace x=0 = 2 + 2(0) / (0^3+a million) = 2 decrease of cos x as x procedures infinity does not exist as cos x oscillates between -a million and a million. -a million <= cos x <= a million -a million/(x^2-sin x) <= cos x /(x^2-sin x) <= a million/(x^2-sin x) as x procedures infinity. the two -a million/(x^2-sin(x)) and a million/(x^2-sin(x)) attitude 0 cos x / (x^2-sin x) is squeezed between 0 and 0. consequently, by using Sandwich Theorem (Squeeze Theorem), cos x /(x^2-sin x) procedures 0. decrease of x^2-4/x-2 as x procedures 2 lim x-->2 (x-2)(x+2)/(x-2) = lim x-->2 (x+2) = 4