Match Functions to Horizontal Asymptotes

[Slimsta]

Homework Statement

"Match functions and their horizontal asymptotes."

Homework Equations

lim f(x)
x-> + infinity

lim f(x)
x-> - infinity

The Attempt at a Solution

http://img198.imageshack.us/img198/7678/45328107.jpg

im just stuck at this... I am sure that everything is right!
but its wrong :(

i was thinking, arctan(x^4-x^2) = tan-1(x^4-x^2)
i saw on a different post, that someone said, graph x^4-x^2 then take the limit and do tan(L) and that's the answer.. but it doesn't seem to work.

the other one that may be wrong (but I am sure its not) is cot-1(x).. i graphed it as 1/tan-1(x) and got 0.6 and -0.6

please help me!

 

[Bohrok]

So your problem was with arctan(x⁴ - x²)?

What is the limit of arctan(x) as x goes to infinity? Does x⁴ - x² also go to infinity as x goes to infinity?

 

[Slimsta]

i don't know... I am assuming so.
thats my question, what wrong with my matching in the picture?

 

[Bohrok]

F is correct. Was there another answer you thought it might be? Do you know why B would not be correct?

Also, what did you mean where you said "but it doesn't seem to work."?

 

[Slimsta]

well the system uses all of the answer i input and shows me right/wrong overall.. it doesn't show me which one is wrong.. so i have to go through each one and check it.
i have check each one 20+ times and also graphed it on a graphing calculator and nothing seems to solve my problem because i keep on getting the same thing..
yeah and B wouldn't be correct because it is only + infinity..
and when plugging in -infinity, x^2 and x^4 make it positive anyways..

how about the cot-1(x) is this one correct?

 

[Bohrok]

cot^-1x ≠ 1/tan^-1x


y = \cot^{-1}x
\cot y = x

\frac{1}{\cot y} = \frac{1}{x}
\tan y = \frac{1}{x}
y = \tan^{-1}\left(\frac{1}{x}\right)
\cot^{-1}x = \tan^{-1}\left(\frac{1}{x}\right)

You should be able to work with this last one now.

 

[Slimsta]

ohhh really?
so that would mean that the answer is 0!
i thought it's 1/tan-1(x)

so the answer would be G=0?

 

[Bohrok]

Yes, looks like that's the only one wrong.

 

[Slimsta]

the system shows that i already had an answer like this.. which means that something else is wrong.

for (6x+1) / (sqrt(3x^2+1))
i get hor.asymptote as pi and -pi..
which means its J.. right?

edit:
so my answers are now in this order:
C
F
B
G
J
A

and looks like everything is right but still the system says its wrong...
maybe that one with J is wrong?

 

[Bohrok]

That one is J (why \pi and -\pi?). Do you know what it's horizontal asymptote is?

 

[Slimsta]

yeah.. horizontal asymptote is the y value that's a line which the graph doesn't pass (the limit of it.. gets really close to it.. and goes to infinity but still doesn't touch it)

if that one is right... then i just don't know.
the first one is the only thing left and its %99.999 right

edit:
the last one, when graphing it, i get 2. when solving it i get 0.. maybe i solved that one wrong?

 

[Bohrok]

if that one is right... then i just don't know.
the first one is the only thing left and its %99.999 right

Which one are you talking about?

 

[Slimsta]

the first matching fraction.. 6x^3+2x+1 / 2x^3-10x^2+13x+100
but i don't think that's the problem..

im more concerned about the last one
f(x)=2-x + sqrt(x^2+1)

im getting 0 when doing it algebraically, but graphing in on a calculator give me a limit of 2

 

[Bohrok]

the first matching fraction.. 6x^3+2x+1 / 2x^3-10x^2+13x+100
but i don't think that's the problem..

That one has a limit of 2, so you're right.

im more concerned about the last one
f(x)=2-x + sqrt(x^2+1)

im getting 0 when doing it algebraically, but graphing in on a calculator give me a limit of 2

Some, like this one, need some creative rewriting so you can find its limit.

 

[Slimsta]

That one has a limit of 2, so you're right.



Some, like this one, need some creative rewriting so you can find its limit.

you meant 3 for the first one right?

for f(x)=2-x + sqrt(x^2+1),
what i did is multiplying both top and bottom by -2+x + sqrt(x^2+1)
which led to:
(4x-3) / (sqrt(x^2+1) +x -2)
from there, i divide both top and bottom by x^2..

(4x-3)/x^2 / (sqrt(x^2+1) +x -2)/x^2

and i got (4/x - 3/x ) / (sqrt(1+ 1/x^2) + x -2)
if x-> infinity..

the top will be immediately 0 and the bottom is 1 which mean 0/1 = 0...
right?

 

[Bohrok]

3, you're right.

You can rewrite it a little easier so you don't make any mistakes, which I think you did

\lim_{x\rightarrow\infty} (2 + \sqrt{x^2 + 1} - x) = 2 + \lim_{x\rightarrow\infty}(\sqrt{x^2 + 1} - x)\left(\frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1} + x}\right)

 

[Slimsta]

oh man... that's what i did wrong?!

now i get (1/0) / 2, which means, doesn't exist

which means it would be "none of the above" for the final answer, am i correct?

 

[Bohrok]

No, it does exist as you saw on your calculator.

 

[Bohrok]

2 + \lim_{x\rightarrow\infty}(\sqrt{x^2 + 1} - x)\left(\frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1} + x}\right) = 2 + \lim_{x\rightarrow\infty}\frac{x^2 + 1 - x^2}{\sqrt{x^2(1 + \frac{1}{x^2})} + x} = 2 + \lim_{x\rightarrow\infty}\frac{1}{x\sqrt{1 + \frac{1}{x^2}} + x}

Can you take it from there?

 

[Slimsta]

No, it does exist as you saw on your calculator.

i got,
(x^2 + 1 - x^2) / (sqrt(x^2 +1) +x) =
1/ (sqrt(x^2 +1) +x)

now i divided both top and bottom by x

1/x / (sqrt(x^2 +1) +x) / x

and got:
1/x / (sqrt(1+ 1/x^2) +1)

>> 1/x^2 = 0 so (sqrt(1+ 0) +1) = 2

top: 1/x = 1/infintiy = 0
0/2 = 0

2+0 = 2
oh man.. so that still 2

then what is wrong??
we went over every single one and still its wrong..

 

[Bohrok]

That one is J (again, do you know what the exact limit of that one is?).

Do you know how many you got wrong?

 

[Slimsta]

That one is J (again, do you know what the exact limit of that one is?).

Do you know how many you got wrong?

6/sqrt(3) = 3.46.. so its close to pi...
i just multiplied top and bottom by sqrt(3x^2 + 1) and solved it..

and no i don't know how many i got wrong.
the system shows wrong overall.. whichever is wrong, that my job to find i guess..

so i have no clue what can be wrong..
we went over every single one in details and still it's wrong. :(

 

[Bohrok]

I'd say all your answers are correct
C
F
B
G
J
A

The system checking your answers has to be wrong,

 

[Slimsta]

I'd say all your answers are correct
C
F
B
G
J
A

The system checking your answers has to be wrong,

omg.. I am in stress now! it's due tomorrow and i have no idea what to do :(

 

[Bohrok]

Someone may have made a simple mistake, such as making the answer to the second one B. All I can suggest is trying different combinations of answers that are very similar to the answers you know are correct.

 

[Slimsta]

Someone may have made a simple mistake, such as making the answer to the second one B. All I can suggest is trying different combinations of answers that are very similar to the answers you know are correct.

i have this feeling that
6x+1 / sqrt(3x^2+1)
is going to be something else, because it's so close to pi.. 3.46... but then there is no -pi.
whatever.. i give up. i have 2 more tries left so i will guess them both and see what happens.

thanks a lot for your time and huge help!

 

[Slimsta]

i got it!
apparently, cot-1(x) is the same as arccot(x) which means,
http://web.viu.ca/wattsv/math122/Overheads/Section35/overhead13.bmp

and the answer was E.. wow... pweee... on the 7/8 tries i got it! woohooo