Horizontal Asymptote: lnx^2/x^2

[gillgill]

is there a horizontal asymptote for
y=(lnx^2)/(x^2)

i know u take the lim to find h.a
but what is ln infinity/infinity?

 

[Jameson]

I could be wrong, but I think L'Hopitals Rule could work here.

 

[gillgill]

what is L'Hopitals Rule ?

 

[bomba923]

Because your limit produces an indeterminate form (that infinity/infinity), you can differentiate separately the numerator and denominator with respect to x, thus applying L'Hopital's Rule . You see :shy:,

\mathop {\lim }\limits_{x \to \infty } \frac{{\ln x^2 }}{{x^2 }} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left( {\ln x^2 } \right)}}{{\frac{d}{{dx}}\left( {x^2 } \right)}} = \mathop {\lim }\limits_{x \to \infty } \frac{1}{{x^2 }} = 0

Hope this helps

 

[gillgill]

o..icic...thats the same for -infinity right?
so the horizontal asymptote is y=0 ?

 

[bomba923]

Correct .

 

[gillgill]

i see...thanks a lot

 

[amcavoy]

Another, quicker way to do this would be to recognize the functions you are dealing with. An algebraic function will "dominate" a logarithmic function for large values of x. Knowing this means that the denominator will go to infinity quicker than the numerator, thus going to zero.