I've been reading about limits, L'Hospitals rule, and rational expressions. I can see when some things get small very fast as x approaches infinity, like 1/x^2. I can see the reverse too. Even with x's in both the numerator and denominator, x always seems to go to infinity or zero. Now I'm stuck - why can't I find a rational expression that goes to -1? Or any negative number really. I thought something like-[((e^-x)-1)/((-e^-x)-1)] maybe, but now I'm not sure. Any guidance would be greatly appreciated!

Question

I've been reading about limits, L'Hospitals rule, and rational expressions.  I can see when some things get small very fast as x approaches infinity, like 1/x^2.  I can see the reverse too.  Even with x's in both the numerator and denominator, x always seems to go to infinity or zero.  Now I'm stuck - why can't I find a rational expression that goes to -1?  Or any negative number really.  I thought something like-[((e^-x)-1)/((-e^-x)-1)] maybe, but now I'm not sure.  Any guidance would be greatly appreciated!

kc_kennylau · Answer

y=(1/x)-1

faithy · Answer

Cool!  It graphs. The calculator doesn't blow up.  If my form is lim x->infinity f(x)/g(x), would it be "cheating" to tag on the negative one?  I'm not sure if there are restrictions on this.  I'm getting excited...this might be it!

kc_kennylau · Answer

y=(1-x)/x then :)

faithy · Answer

Your answers both work (thank you!!!)...but, the book says this crazy thing.  In the Indeterminate Forms section.  It gives an example lim x->inf lnx/x-1  (close to the formula above) but the text states, "It isn't obvious how to evaluate this limit because both the numerator and denominator become large as x->infinity."  Basically, top wins, limit is infinity, bottom wins, limit is zero.  Then, "Or there may be some compromise, in which case the answer will be some finite positive number."  This is what has me perplexed.  Why does it say positive?  I am totally missing something here.

faithy · Answer

You just showed me something that graphs like it's a negative number!

kc_kennylau · Answer

y=(2x+3)/(x+4) would have 2

kc_kennylau · Answer

y=(4x-9)/(2x+5) would be 2

faithy · Answer

Oh, yeah!  That is that powers rule.  Coefficient of top - coefficient of bottom (for when there is an x on both sides.)  We use the highest power one for this.

kc_kennylau · Answer

y=(9-10x)/(5x+4) would be -2

kc_kennylau · Answer

But the coefficient of x would be negative

kc_kennylau · Answer

which is considered "cheating"

faithy · Answer

Ack!  Why is it cheating?  This is what I don't get!

kc_kennylau · Answer

Because you have so declared. :P

faithy · Answer

lolol  :D  You're awesome.  I'm studying for finals and this has really been distracting...  Wait a minute.  If there is a negative on top, wouldn't that go to negative infinity?

kc_kennylau · Answer

y=(3-x^2)/(6x+2) would go to negative infinity

faithy · Answer

I think I'm starting to get it...  Brute force formula attempts are never my first choice, but a girl has to try.  I seriously think I've tried every combination available.  At least, up until calculus...  Maybe a few more math classes down the road, they'll change all the rules and there WILL be a way to do this.  :D  Thanks for helping me.  A bright spot in my all-nighter.

kc_kennylau · Answer

I'm flattered :)

faithy · Answer

I'm sure I'll be back.  It is, after all, math.  :P  Night, night. Thanks again.

kc_kennylau · Answer

No problem!

faithy · Answer

Update:  I asked my professor about this today after class and putting f(x)/g(x) in parenthesis, then making the whole thing negative, is an illegal math move.  It's no longer a rational expression at that point.  Making the numerator or denominator negative changes the direction of the limit, so it's no longer taking the limit as x->infinity.  Fact:  The lim x->infinity g(x)/f(x) cannot be a negative number!

kc_kennylau · Answer

Oh... I've never thought of that. Thanks! :)