1. ## Verify limit?

Hi All,

I've been wrapping my head around this limit problem and I think my error is stemming from some bad factoring but I could be wrong:

Limit as x approaches 0 from positive infinity
((3^x)-(3^-x))/((3^x)+(3^-x)

The answer should be: approaches 1.

The answer I always get is 0.

Any help would be much appreciated!

2. ## Re: Verify limit?

Originally Posted by jokem
Hi All,

I've been wrapping my head around this limit problem and I think my error is stemming from some bad factoring but I could be wrong:

Limit as x approaches 0 from positive infinity
((3^x)-(3^-x))/((3^x)+(3^-x)

The answer should be: approaches 1.

The answer I always get is 0.

Any help would be much appreciated!
\displaystyle \begin{align*} \frac{3^x - 3^{-x}}{3^x + 3^{-x}} &= \frac{3^x\left( 3^x - 3^{-x}\right)}{3^x \left( 3^x + 3^{-x} \right)} \\ &= \frac{3^{2x} - 1}{3^{2x} + 1} \\ &= 1 - \frac{2}{3^{2x} + 1} \end{align*}

I agree that this should approach 0 as you make x approach 0.

3. ## Re: Verify limit?

You don't even need to take limits. This function is continuous on all of $\mathbb{R}$.

f(0) does equal 0.

But the limit at infinity is 1... maybe the problem was read or interpreted wrong?

4. ## Re: Verify limit?

Wow, thank you very much!