I'm completely stuck on the limit of sinxcos(1/x) as x goes to 0. If you are allowed to take the arcsin of epsilon, then I should be fine, but I have a feeling you aren't allowed to do that. Any hints?
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Originally Posted by jsmith90210 I'm completely stuck on the limit of sinxcos(1/x) as x goes to 0. Surely you can see that $\displaystyle \left| {\sin (x)\cos \left( {\frac{1}{x}} \right)} \right| \le \left| {\sin (x)} \right|.$
yes, I can see that, but then I have abs(sinx)<epsilon, i need to get from there to abs(x)<some variation of epsilon and I don't know how to do that
Originally Posted by jsmith90210 yes, I can see that, but then I have abs(sinx)<epsilon, i need to get from there to abs(x)<some variation of epsilon and I don't know how to do that Do you know that $\displaystyle |\sin(x)|\le |x|~?$
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