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Please help me!! Ahh calculus
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\[\lim_{x \rightarrow 0} \left[ x ^{2}\sin (1/x) \right]\]
No limit. You can't divide by 0.
i tried that. it doesn't work like that my teacher said I was wrong
Does not exist.
Answer is 0 Use the squeeze theroem
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what is the squeeze theroem
It says that if you have three functions, and the middle function lies in-between the upper and lower one, the limit of the middle function is above the lower and below the upper function. I didn't really understand it, seemed like common sense. Maybe someone can explain more fully to you.
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