Is the Approximation tanh(t) ~ t - t^3/3 + 5t^5 Accurate as t Approaches Zero?

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SUMMARY

The approximation tanh(t) ~ t - t^3/3 + 5t^5 as t approaches 0 is confirmed as accurate. The limit of tanh(t) divided by the approximation approaches 1 as t approaches 0, validating the statement. To analyze this, one should start with the unapproximated representation of tanh(t) using the limit of sinh(t)/cosh(t). Applying L'Hôpital's rule may be necessary if the limit results in an indeterminate form.

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sam_0017
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so is the statement is true ??

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 ??
 
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sam_0017 said:
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 ??

Hey sam_0017 and welcome to the forums.

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'hospital's 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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