What is #lim_(x->0) ln(sin x + cos x)/x# ?
1 Answer
Mar 9, 2017
Explanation:
Intuitive analysis
Notice that for small values of
#sin x ~~ x#
#cos x ~~ 1#
#ln (x+1) ~~ x#
So:
#ln(sin x+cos x) ~~ ln(x+1) ~~ x#
So we should expect:
#lim_(x->0) ln(sin x + cos x)/x = 1#
Semi-formal analysis
#sin x = sum_(n=0)^oo ((-1)^n)/((2n+1)!) x^(2n+1) = x+O(x^3)#
#cos x = sum_(n=0)^oo ((-1)^n)/((2n)!) x^(2n) = 1+O(x^2)#
#ln (x + 1) = sum_(n=0)^oo ((-1)^n)/(n+1) x^(n+1) = x+O(x^2)#
So:
#ln(sin x + cos x)/x = ln(x+1+O(x^2))/x#
#color(white)(ln(sin x + cos x)/x) = (x+O(x^2))/x#
#color(white)(ln(sin x + cos x)/x) = 1+O(x) -> 1" "# as#x -> 0#