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lim x -> 0 (x-1)/(3sin(x) - cos(x))
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evaluate the limit of the trigonomoteric limit
(0-1)/(3sin(0)-cos(0)) =-1/(3*0-1) =-1/(0-1) =-1/(-1) =1
we cant substitute first
the function is continuous at x=0 so yes we can
:s ok i guess
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so thats not the same as lim x-> 0 2sin (3x)/ 5x which would be equal to 6/5?
this function is discontinuous at x=0 we can manipulate this function though to see what happens as x->0 remember sinx/x->1 as x->0 so 3/3*2sin(3x)/5x=6/5*sin(3x)/3x=6/5*(1) since sin(3x)/3x->1 as x->0
ok thank you i get it
i forgot about continuity
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