Help!! Calculus problem!!! help meeee!!!

[Omegachi]

Ahhh....i have a calculus final tomorrow!!! and i don't understand a problem 🙁

how do you find the horizontal asymptote of f(x)=(1-x)/(x+2) ??

i know the vertical asym. is -2 cause thats what make it undefined...

whats the horizontal one??!! ahhhhhhhhh. help

 

[minendo]

dont you just set f(x) = 0

 

[minendo]

im an idiot. no wonder i did so poorly in calculus

 

[Omegachi]

okay....nevermind, I figured it out. thanks anyways

i just take the limit as x->infinite, the answer should be -1

 

[IronChefM]

try taking the limit of the fxn as x --> infinity
use l'hopital if you need to

what does your book say?

 

[HonkeyDonk]

mimendo: setting f(x) = 0 will only find the x-intercept.

To find the horizontal asymptote, I would just plug inifinity for x.

the top would become -infinity, and the bottom would be +infinity, and i Know you can't devide infinity/infinity (techincally, it's infinity) but, it would end up being -1.

horz. asym. = -1

another way is to find the highest powers of x for the top and bottom and divide accordingly.

For here, we have -x^1 on top, and x^1 on top, so take out anything that's not the highest power of x and that leaves you with -x^1/x^1 = -1.

 

[LoqT]

The horizontal asymptote is found by getting the limit of f(x) as x approaches infinity. (Also what number does f(x) approach as x goes to infinity. The 1 and the 2 in the problem become irrelevant because they affect the result less and less as x gets larger, therefore the answer would be -1.

 

[DuffmanOhYeah]

As we do a linear degredation of the cotangent in question, it is absolutely imperative to know the vector coordinates of one's bellybutton.

 

[pillage2001]

OH god, brain is dead......Help me...you guys are CRAZY.......

 

[DuffmanOhYeah]

<< OH god, brain is dead......Help me...you guys are CRAZY....... >>


On the contrary, Im not crazy at all. My answer was rooted in the very foundation of mathematical principle. 😛

 

[OUdejavu]

(1-x)/(x+2) = 1/(x+2) - x/(x+2) = 1/(x+2) - 1/(1+(2/x)) as x->inf we get 0 - 1/(1+0) = -1