Limit of function

  • #1
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1
Member warned about not using the homework template
Hi,

I'm trying to find the limit as x tends to zero of the function (tanhx-x)/x
this is what i have but i have no idea if i am on the right lines?
lim x-> 0 can be split up in to two problems:
limx->0 (tanhx)/x - limx->0 x/x
limx->0 x/x = 1
limx->0 (tanhx)/x can be expressed as
limx->0 ((e^x-e^-x)/e^x+e^-x))/x
=
limx->0 (x(e^x-e^-x)/e^x+e^-x)) =0
which leave the limit as -1 but wolfram gives a limit of 0

how should i be approaching this problem?
many thanks
Ryan
 

Answers and Replies

  • #2
BiGyElLoWhAt
Gold Member
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in your final expression, I'm not sure how you're getting -1
[e^0 - e^(-0)] =? I think you plugged in your limit wrongly.

Edit: Oh wait, I thought you were saying that the limit of tanh was -1.
Your problem lies in the fact that ##\frac{tanh(x)}{x} \neq \frac{x(e^x-e^{-x}}{e^x +e^{-x})}##

##\frac{tanh(x)}{x} = \frac{e^x-e^{-x}}{x(e^x +e^{-x})}##
 
Last edited:
  • #3
33
1
With the x/x you can use l'hopitals rule to get round the fact that you have somepthing that is tending to 0/0 but for the tanh(x)/x part you cant? so how do you evaluate it? and im a bit confused as to how
tanh(x)x=ex−e(−xx(ex+e(−x))
Because i thought
((e^x-e^-x)/e^x+e^-x))/x was the same as:
((e^x-e^-x)/e^x+e^-x)) times x/1 ?


many thanks for your reply :)
Ryan
 
  • #4
BiGyElLoWhAt
Gold Member
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You can use the hospital rule. But you didn't express tanh x/x correctly.
##\frac{tanh x}{x} = \frac{1}{x}tanh x = \frac{1}{x}\frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{e^x-e^{-x}}{x(e^x + e^{-x})}##
 
  • #5
BiGyElLoWhAt
Gold Member
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In your case, you end up with a 0/1, but in this case you don't.
 
  • #6
33
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ah brilliant much clearer thanks! :)
 
  • #7
BiGyElLoWhAt
Gold Member
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Not a problem.
 

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