Expert: Paul Klarreich - 1/3/2010

Question
What are the vertical and horizontal asymptotes of this function?
... f(x)= [ln(x+2)]/[ln(X^2+3x+3)]...... ....
my answer is :
one vertical asymptote which is x=-2  ......
and one horizontal asymptote which is y=0.5  ........

Answer
Questioner: Jaim
Country: Malaysia
Category: Calculus
Private: No
Subject: Asymptotes...
Question: What are the vertical and horizontal asymptotes of this function?
... f(x)= [ln(x+2)]/[ln(X^2+3x+3)]...... ....
my answer is :
one vertical asymptote which is x=-2  ......
and one horizontal asymptote which is y=0.5  ........
.............................................................

Well, if you already have the answers, I'll just take a nap.

Yes, the function is undefined for x <= -2, and as  x -> -2,

x+2 --> 0,  so ln(x+2) --> - infinity.

And at x = -2, X^2+3x+3 = 1, but when  -2 < x < -1, that is negative, so your fraction will approach PLUS infinity.

So far so good.

y = 0.5 looks OK for the h.a.

Do this:

  ln(x+2)
------------- =
ln(x^2+3x+3)


   ln(x(1+2/x)
------------------------- =
ln(x^2(1 + 3/x + 3/x^2)


   ln x + ln (1+2/x)
------------------------------- =
ln (x^2) + ln(1 + 3/x + 3/x^2)

and as  x -> infinity, this -->

   ln x + ln (1+0)
-------------------------- =
ln (x^2) + ln(1 + 0 + 0)

   ln x + ln (1)
------------------- =
ln (x^2) + ln(1)

   ln x
----------- =
ln (x^2)

 ln x
------- = 1/2 !!!
2 ln x

That's it.