How do you use the squeeze law of limits to find a limit involving a third root?

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The discussion revolves around using the squeeze law of limits to evaluate the limit of x^2cos(1/third root of x) as x approaches 0. Participants clarify that the cosine function oscillates between -1 and 1, which helps establish bounds for the product x^2cos(1/third root of x). By recognizing that the limits of x^2 and -x^2 both approach 0 as x approaches 0, the squeeze theorem can be applied effectively. The importance of ensuring that the limits of the bounding functions are the same is emphasized. Ultimately, the limit can be determined to be 0.
Monsu
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i am quite confused with that law, would someone pls just explain it ina few words?
well, the actual question goes like this:
Use the squeez law of limits to find lim x->0 x^2cos(1/thirdrootx).

(third root of 8 = 2 , 2x2x2 = 8, in case u do not understand what i mean by third root, i can't type the sign) :confused:
 
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well if you graph cos you will notice it oscillates (sp?) between -1 and 1, so if you multiply cos(angle) with x^2 it will have a "amplitud", new max values between x^2 and -x^2.

so if you know the limit of x^2 and -x^2 you will know where does cos(angle) goes. I forgot and if the limits of both are the same!, very important.
 
so whatever happens to the 1/thirdrootX ??
 
Monsu said:
so whatever happens to the 1/thirdrootX ??

Try many values in the calculator of any form with cosine and see what you observe.
 
Got it! thanks.
 

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