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Answer 1 · 2017-06-05T17:57:23

#lim_{x\to\infty} x-xcos(1/x)= 0#

Say #"f"(x)=x-xcos(1/x)#.

Using the series #cos(x)=1-x^2/(2!)+x^4/(4!)+...#.

Then,

#"f"(x)=x-x(1-1/(2!x^2)+1/(4!x^4)+...)#,
#"f"(x)=1/(2!x)-1/(4!x^3)+1/(5!x^4)-...#.

As #x\to\infty#, #1/x^n \to 0# for #n>0#.

Then, #"f"(x) \to 0# as #x\to\infty#.

Answer 2 · 2017-06-06T14:26:26

#lim_(xrarroo)(x-xcos(1/x)) = 0# by factoring and a fundamental trigonometric limit.

Recall that #lim_(hrarr0)(1-cos h)/h = 0#

#lim_(xrarroo)(x-xcos(1/x)) = lim_(xrarroo)(x[1-cos(1/x)]) #

# = lim_(xrarroo)(1-cos(1/x))/(1/x)#

Now as #xrarroo#, we have #1/x rarr 0#.

So, with #h = 1/x#, this limit is

#lim_(hrarr0)(1-cos h)/h = 0#