# Homework Help: Help With Infinite Exponential Limit

1. Aug 29, 2007

### alba_ei

1. The problem statement, all variables and given/known data
$$\lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}}$$
2. Relevant equations
$$\lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}} = - 1$$

3. 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.

$$\lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}}$$

$$= \lim_{x \to - \infty} \frac{3^{-x}(3^{2x}-1)}{3^{-x}(3^{2x}+1)}$$

$$= \lim_{x \to - \infty} \frac{3^{2x}-1}{3^{2x}+1}$$

$$= \lim_{x \to - \infty} \frac{1-\frac{1}{3^{2x}}}{1+\frac{1}{3^{2x}}}$$

$$= \lim_{x \to - \infty} \frac{1}{1} = 1$$
I got 1, not -1

Last edited: Aug 29, 2007
2. Aug 29, 2007

### EnumaElish

How did you get +1?

3. Aug 29, 2007

### 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.

4. Aug 29, 2007

### cristo

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

5. Aug 29, 2007

### 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 $3^{2x}=e^{2x\ln{3}}$

Last edited: Aug 29, 2007
6. Aug 29, 2007

### EnumaElish

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

7. Aug 29, 2007

### cristo

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

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.