Compute the limit x→∞ [e^x + e^(-x)] / [e^x - e^(-x)]

[hndalama]

lim _x→∞ [e^x + e^-x] / [e^x - e^-x]

LHopital's rule doesn't seem to work. i still end up with ∞/∞

 

[pka]

lim _x→∞ [e^x + e^-x] / [e^x - e^-x]
LHopital's rule doesn't seem to work. i still end up with ∞/∞

$\displaystyle \dfrac{e^x+e^{-x}}{e^x-e^{-x}}=\dfrac{e^{2x}+1}{e^{2x}-1}$ now use LHopital's rule.

 

[hndalama]

$\displaystyle \dfrac{e^x+e^{-x}}{e^x-e^{-x}}=\dfrac{e^{2x}+1}{e^{2x}-1}$ now use LHopital's rule.

Oh thank you

 

[ksdhart2]

Or, another way you could have tackled this problem is to use hyperbolic sine and cosine. Recall that they are defined thusly:

$\displaystyle sinh(x)=\dfrac{e^x-e^{-x}}{2}$ and $\displaystyle \dfrac{cosh(x)=e^x+e^{-x}}{2}$

From these definitions it follows that:

$\displaystyle 2sinh(x)=e^x-e^{-x}$ and $\displaystyle 2cosh(x)=e^x+e^{-x}$

So your limit is then:

$\displaystyle \displaystyle \lim _{x\to \infty }\left(\frac{e^x+e^{-x}}{e^x-e^{-x}}\right) = \lim _{x\to \infty }\left(\frac{2cosh(x)}{2sinh(x)}\right) = \lim _{x\to \infty }\left(\frac{cosh(x)}{sinh(x)}\right)$

Since hyperbolic sine and cosine obey many of the same relationships as do regular sine and cosine, we can see that the above is equal to the hyperbolic cotangent. And that limit has a known value, namely 1.

Incidentally, the nature of hyperbolic sine and cosine is why L'Hopital's rule wouldn't work without rewriting the equation: The derivative of sinh(x) is cosh(x) and the derivative of cosh(x) is sinh(x). They don't even oscillate negative signs, like regular sine and cosine do. So all L'Hopital's Rule was doing was switching between coth(x) and tanh(x).