limits toward neg. infinity

**Evan.Kimia** · December 17th 2009, 12:54 PM

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** · December 17th 2009, 12:59 PM

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** · December 17th 2009, 01:19 PM

Ah, makes perfect sense. Thank you.

**Raoh** · December 17th 2009, 01:41 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** · December 17th 2009, 02:56 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.

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