How do you use the Squeeze Theorem to find $lim sqrt(x^3+x^2) * sin (pi/x)$ as x approaches zero?

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Answer 1 · 2015-10-07T01:36:52

See the explanation below.

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

For all $x != 0$ for which the square root is real, $sqrt(x^3+x^2) >0$ , so we can multiply the inequality without changing the direction.

$-sqrt(x^3+x^2) <= sqrt(x^3+x^2)sin(pi/x) <= sqrt(x^3+x^2)$ .

We observe that $lim_(xrarr0)-sqrt(x^3+x^2) = -sqrt(0+0) = 0$ ,

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

So, by the Squeeze Theorem,

$lim_(xrarr0)sqrt(x^3+x^2)sin(pi/x) = 0$ .

How do you use the Squeeze Theorem to find lim sqrt(x^3+x^2) * sin (pi/x) lim sqrt(x^3+x^2) * sin (pi/x) as x approaches zero?