Question #7e404

2 Answers
Jun 5, 2017

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

Explanation:

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.

Jun 6, 2017

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

Explanation:

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