
#1
Mar2912, 04:34 AM

P: 24

tanh(t)~t t^3/3 + 5t^5 as t[itex]\rightarrow[/itex] 0
i find that the lim tanh (t)/(t t^3/3 + 5t^5) = 1 when t[itex]\rightarrow[/itex] 0. so is the statement is true ?? 



#2
Mar2912, 04:48 AM

P: 4,570

If you want to find a limit of something you start out with the unapproximated representation. So it would be wise to find lim (x>0) tanh(x) = lim(x>0)sinh(x)/cosh(x). If you get something like infinity/infinity then you use things like l'hopitals rule, but if you get a determinate form then that's your answer. Also remember that sinh(x) and cosh(x) are continuous (actually tanh(x) is as well). You should get an answer of 0, but it would probably be beneficial for you if working was shown in this thread. 


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