How do you find the limit of $x^{2} cos (\frac{π}{x})$ $x^2cos(pi / x)$ as x approaches 0?

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How do you find the limit of $x^{2} cos (\frac{π}{x})$ $x^2cos(pi / x)$ as x approaches 0?

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Answer 1 · 2015-05-05T20:35:21

cos(1/x) oscillates between 1 and -1 really fast when aproaching to 0 but it always is a finite number. If you are multipliying this oscillating function to x^2 which modullates the oscillation it will oscillate between x^2 and -x^2.
So if you make the limit going to 0 it will "oscillating" between +0 and -0 which you see is 0.

Answer 2 · 2015-05-06T01:44:18

The limit is $0$ $0$ .

To show this, use the squeeze (pinch, sandwich) theorem.

$- 1 \leq cos (\frac{π}{x}) \leq 1$ $-1<= cos (pi/x)<= 1$ for $x \neq 0$ $x!=0$

Because $x^{2} > 0$ $x^2 >0$ for all $x \neq 0$ $x!=0$ , we can multiply the inequality without reversing it:

$- x^{2} \leq x^{2} cos (\frac{π}{x}) \leq x^{2}$ $-x^2<= x^2cos (pi/x)<= x^2$ for $x \neq 0$ $x!=0$ .

Note that

$lim_(xrarr0)(-x^2) in=0=lim_(xrarr0)x^2$ ,

So the squeeze theorem (or whatever you call it) tells us that

$lim_(xrarr0) x^2cos (pi/x) =0$

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How do you find the limit of $x^{2} cos (\frac{π}{x})$ $x^2cos(pi / x)$ as x approaches 0?

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