Help With Infinite Exponential Limit

[alba_ei]

Homework Statement

<br /> \lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}}<br />

Homework Equations

<br /> \lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}} = - 1<br />

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.

<br /> \lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}}<br />

<br /> = \lim_{x \to - \infty} \frac{3^{-x}(3^{2x}-1)}{3^{-x}(3^{2x}+1)}<br />

<br /> = \lim_{x \to - \infty} \frac{3^{2x}-1}{3^{2x}+1}<br />

<br /> = \lim_{x \to - \infty} \frac{1-\frac{1}{3^{2x}}}{1+\frac{1}{3^{2x}}}<br />

<br /> = \lim_{x \to - \infty} \frac{1}{1} = 1<br />
I got 1, not -1

 

[EnumaElish]

How did you get +1?

 

[ELESSAR TELKONT]

you can do an algebraic trick to get that \frac{3^{x}-3^{-x}}{3^{x}+3^{-x}}=\frac{3^{2x}-1}{3^{2x}+1} and then the limit is trivial.

 

[cristo]

I'm not sure how you got from the left hand side to the middle of your solution. Could you explain your steps further?

 

[ELESSAR TELKONT]

remember that when you take the limit of an exponential when the variable tends to -\infty you get that this limit is 0. And remember that 3^{2x}=e^{2x\ln{3}}

 

[EnumaElish]

Does 1/3^{2x} ---> 0 as x ---> -Infinity?

 

[cristo]

I think I messed up my reply by not quoting anyone, and now you've edited and put in more of your solution! Anyway...

Homework Statement

<br /> \lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}}<br />

Homework Equations

<br /> \lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}} = - 1<br />

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.

<br /> \lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}}<br />

<br /> = \lim_{x \to - \infty} \frac{3^{-x}(3^{2x}-1)}{3^{-x}(3^{2x}+1)}<br />

<br /> = \lim_{x \to - \infty} \frac{3^{2x}-1}{3^{2x}+1}<br />

<br /> = \lim_{x \to - \infty} \frac{1-\frac{1}{3^{2x}}}{1+\frac{1}{3^{2x}}}<br />

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.

<br /> = \lim_{x \to - \infty} \frac{1}{1} = 1<br />
I got 1, not -1