Easy yet tricky limit question (showing my work)

I have to show my work on a problem: $\displaystyle lim_{x \to -\infty}\frac{e^x-e^{-x}}{e^x+e^{-x}}$

Ok, so first thing: divide by fastest growing term in the denominator, which in this case as x approaches $\displaystyle -\infty$ is actually: $\displaystyle e^{-x}$

So: $\displaystyle lim_{x \to -\infty}\frac{\frac{e^x}{e^{-x}}-\frac{e^{-x}}{e^{-x}}}{\frac{e^x}{e^{-x}}+\frac{e^{-x}}{e^{-x}}}$ = $\displaystyle lim_{x \to -\infty}\frac{e^{2x}-1}{e^{2x}+1}$

This is where I start to get a little confused. How exactly do I mathematically express that as x approaches $\displaystyle -\infty$ $\displaystyle e^{2x}$ approaches zero?

I can say: $\displaystyle \frac{lim_{x \to -\infty}e^{2x}-1}{lim_{x \to -\infty}e^{2x}+1}$, but then what? I have seen direct substitution with $\displaystyle -\infty$ but isn't this bad form? You can't do arithmetic on infinity.

So then does it just "jump" to: $\displaystyle \frac{0-1}{0+1} = \frac{-1}{1} = -1$?

Or is there some merit to doing this next (after splitting the func. to a quotient of limits): $\displaystyle \frac{e^{2*-\infty}-1}{e^{2*-\infty}+1} = \frac{e^{-\infty}-1}{e^{-\infty}+1}$? I thought this had no meaning; if it does what is the next step?

Thank you for any help, I know I got the answer, but it's the showing of the steps I am a little tripped up on. IF there is anything missing, can you also explain then how I can exactly express the other horizontal limit $\displaystyle lim_{x \to \infty}\frac{e^x-e^{-x}}{e^x+e^{-x}}$?

Re: Easy yet tricky limit question (showing my work)

Quote:

Originally Posted by

**rortiz81** I have to show my work on a problem: $\displaystyle lim_{x \to -\infty}\frac{e^x-e^{-x}}{e^x+e^{-x}}$

Ok, so first thing: divide by fastest growing term in the denominator, which in this case as x approaches $\displaystyle -\infty$ is actually: $\displaystyle e^{-x}$

So: $\displaystyle lim_{x \to -\infty}\frac{\frac{e^x}{e^{-x}}-\frac{e^{-x}}{e^{-x}}}{\frac{e^x}{e^{-x}}+\frac{e^{-x}}{e^{-x}}}$ = $\displaystyle lim_{x \to -\infty}\frac{e^{2x}-1}{e^{2x}+1}$

This is where I start to get a little confused. How exactly do I mathematically express that as x approaches $\displaystyle -\infty$ $\displaystyle e^{2x}$ approaches zero?

I can say: $\displaystyle \frac{lim_{x \to -\infty}e^{2x}-1}{lim_{x \to -\infty}e^{2x}+1}$, but then what? I have seen direct substitution with $\displaystyle -\infty$ but isn't this bad form? You can't do arithmetic on infinity.

No, you shouldn't just replace x with "infinity" but you can use the basis limit "properties":

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

And now you can replace those **limits** with their values to get $\displaystyle \frac{0- 1}{0+ 1}$.

Quote:

So then does it just "jump" to: $\displaystyle \frac{0-1}{0+1} = \frac{-1}{1} = -1$?

Or is there some merit to doing this next (after splitting the func. to a quotient of limits): $\displaystyle \frac{e^{2*-\infty}-1}{e^{2*-\infty}+1} = \frac{e^{-\infty}-1}{e^{-\infty}+1}$? I thought this had no meaning; if it does what is the next step?

Thank you for any help, I know I got the answer, but it's the showing of the steps I am a little tripped up on. IF there is anything missing, can you also explain then how I can exactly express the other horizontal limit $\displaystyle lim_{x \to \infty}\frac{e^x-e^{-x}}{e^x+e^{-x}}$?

Re: Easy yet tricky limit question (showing my work)

Thank you, I skipped the sum/difference of limits law in both the numerator and denominator for brevity, but I will include that in my work.