Help With Infinite Exponential Limit

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alba_ei
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Homework Statement


[tex] \lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}}[/tex]

Homework Equations


[tex] \lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}} = - 1[/tex]

The Attempt at a Solution


Im getting trouble when I try to evaluate this limit, altough the answer is -1 idont know how to get to it.

[tex] \lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}}[/tex]

[tex] = \lim_{x \to - \infty} \frac{3^{-x}(3^{2x}-1)}{3^{-x}(3^{2x}+1)}[/tex]

[tex] = \lim_{x \to - \infty} \frac{3^{2x}-1}{3^{2x}+1}[/tex]

[tex] = \lim_{x \to - \infty} \frac{1-\frac{1}{3^{2x}}}{1+\frac{1}{3^{2x}}}[/tex]

[tex] = \lim_{x \to - \infty} \frac{1}{1} = 1[/tex]
I got 1, not -1
 
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on Phys.org
you can do an algebraic trick to get that [itex]\frac{3^{x}-3^{-x}}{3^{x}+3^{-x}}=\frac{3^{2x}-1}{3^{2x}+1}[/itex] and then the limit is trivial.
 
remember that when you take the limit of an exponential when the variable tends to [itex]-\infty[/itex] you get that this limit is 0. And remember that [itex]3^{2x}=e^{2x\ln{3}}[/itex]
 
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I think I messed up my reply by not quoting anyone, and now you've edited and put in more of your solution! Anyway...

alba_ei said:

Homework Statement


[tex] \lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}}[/tex]

Homework Equations


[tex] \lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}} = - 1[/tex]

The Attempt at a Solution


Im getting trouble when I try to evaluate this limit, altough the answer is -1 idont know how to get to it.

[tex] \lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}}[/tex]

[tex] = \lim_{x \to - \infty} \frac{3^{-x}(3^{2x}-1)}{3^{-x}(3^{2x}+1)}[/tex]

[tex] = \lim_{x \to - \infty} \frac{3^{2x}-1}{3^{2x}+1}[/tex]

[tex] = \lim_{x \to - \infty} \frac{1-\frac{1}{3^{2x}}}{1+\frac{1}{3^{2x}}}[/tex]
There is no need to do this step. You should note that, since we are taking the limit x tends to negative infty, 3^2x tends to zero.
[tex] = \lim_{x \to - \infty} \frac{1}{1} = 1[/tex]
I got 1, not -1