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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

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