Friday, August 24, 2018

calculus - Solve limxto0fracsinxxx3





I'm trying to solve this limit




limx0sinxxx3




Solving using L'hopital rule, we have:



limx0sinxxx3=limx0cosx13x2=limx0sinx6x=limx0cosx6=16.




Am I right?



I'm trying to solve this using change of variables, I need help.



Thanks



EDIT



I didn't understand the answer and the commentaries, I'm looking for an answer using change of variables.


Answer




I suppose the below counts as a change of variable.



Assuming that the limit exists, then you can compute the limit as follows:



Replace x by 3x, then the limit (say L) is



L=limx0sin3x3x27x3=limx03sinx3x4sin3x27x3=


limx019(sinxxx3)limx0427(sin3xx3)



(we used the formula sin3x=3sinx4sin3x).




Thus we get



L=L9427L=16



Of course, we still need to prove that the limit exists.


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