# Thread: asymptotes of ln

1. ## asymptotes of ln

Find all vertical and horizontal asymptotes of

f(x) = ln(2x)/ln(x^2)

I understand for vertical asymptote, there is one at x = 1, because denominator becomes 0.

But for horizontal asymptote, i know as x -> inf, the denominator will increase faster than the numerator because of the squared, but is there a way to show it mathematically?

Like with polynomials, you divide by the highest power, but with ln, what do you do?

Thanks.

2. Originally Posted by ywd
Find all vertical and horizontal asymptotes of

f(x) = ln(2x)/ln(x^2)

I understand for vertical asymptote, there is one at x = 1, because denominator becomes 0.

But for horizontal asymptote, i know as x -> inf, the denominator will increase faster than the numerator because of the squared, but is there a way to show it mathematically?

Like with polynomials, you divide by the highest power, but with ln, what do you do?

Thanks.
note that $\displaystyle \ln 2x = \ln 2 + \ln x$

and

$\displaystyle \ln (x^2) = 2 \ln |x|$

3. oh i see

so its limit of

(ln 2 + ln x)/(2ln x)

=> ln2/2lnx + lnx/2lnx

=> 0 + 1/2

=> 1/2

:$can you tell me if i did it right? we didn't cover this in our lecture >.< 4. Originally Posted by ywd oh i see so its limit of (ln 2 + ln x)/(2ln x) => ln2/2lnx + lnx/2lnx => 0 + 1/2 => 1/2 :$ can you tell me if i did it right? we didn't cover this in our lecture >.<
yes.

we only had to take the limit going to infinity here, since the limit going to -infinity doesn't exist. thus, ln(|x|) = ln(x) for positive values, and so, the ln(x)'s cancelled, leaving you with what you have