Thread: limits toward neg. infinity

I understand that in order to solve, i would divide by the highest power of x. But what is he doing here that gives 7/7 for x > -infinity?

Originally Posted by Evan.Kimia

I understand that in order to solve, i would divide by the highest power of x. But what is he doing here that gives 7/7 for x > -infinity?

$\displaystyle \lim_{x \rightarrow - \infty } \exp(x)=0$

Ah, makes perfect sense. Thank you.

if i'm not mistaking your b question doesn't look right.
$\displaystyle \lim_{x \rightarrow \infty } \frac{7-\exp x }{7+8\exp x } $= $\displaystyle \lim_{x \rightarrow \infty } \frac{\exp x( \frac{7 }{\exp x }-1) }{\exp x( \frac{ 7}{\exp x }+8) }$ = $\displaystyle \lim_{x \rightarrow \infty } \frac{( \frac{7 }{\exp x }-1) }{( \frac{ 7}{\exp x }+8) }$$\displaystyle = -\frac{ 1}{8 } $

I'm sure for A he's using the fact that both $\displaystyle -e^{x}$ and $\displaystyle 8e^{x}$ towards negative infinity are both zero and that renders....

$\displaystyle \frac{7 - 0}{7 + 0} $

and B is correct.

LOL. whoops, just saw the previous post.