limits toward neg. infinity

**Evan.Kimia** · Dec 17th 2009, 11:54 AM

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?

**Raoh** · Dec 17th 2009, 11:59 AM

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?

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

**Evan.Kimia** · Dec 17th 2009, 12:19 PM

Ah, makes perfect sense. Thank you.

**Raoh** · Dec 17th 2009, 12:41 PM

if i'm not mistaking your b question doesn't look right.

$\lim_{x \rightarrow \infty } \frac{7-\exp x }{7+8\exp x }$ = $\lim_{x \rightarrow \infty } \frac{\exp x( \frac{7 }{\exp x }-1) }{\exp x( \frac{ 7}{\exp x }+8) }$ = $\lim_{x \rightarrow \infty } \frac{( \frac{7 }{\exp x }-1) }{( \frac{ 7}{\exp x }+8) }$ $= -\frac{ 1}{8 }$

**JSB1917** · Dec 17th 2009, 01:56 PM

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

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

and B is correct.

LOL. whoops, just saw the previous post.

Thread: limits toward neg. infinity

LinkBack

Thread Tools

Display

limits toward neg. infinity

Similar Math Help Forum Discussions

Limits at infinity

limits to infinity

Limits at Infinity

limits to infinity

Limits at Infinity

Search Tags