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Question #8cc23

1 Answer

Nov 30, 2016

Please see below.

Explanation:

$-1 <= sin(pi/x) <= 1$ for all $x != 0$ .

$sqrt(x^3+x^2) > 0$ so we can multiply without changing the inequalities.

$-sqrt(x^3+x^2) <= sqrt(x^3+x^2)sin(pi/x) <= sqrt(x^3+x^2)$ $" "$ for all $x != 0$ .

Observe that $lim_(xrarr0)-sqrt(x^3+x^2) = 0$ and $lim_(xrarr0)sqrt(x^3+x^2) = 0$ .

Therefore, by the squeeze theorem, $lim_(xrarr0)sqrt(x^3+x^2)sin(pi/x) = 0$

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