Square roots & oblique asymptotes (1 Viewer)

[ssglain]

I'm just a bit curious..
For f(x) which has an oblique asymptote of A(x), does the asymptote of g(x) = SQRT[f(x)] become SQRT[A(x)]? Provided, of course, that as x->infinity f(x)>0 & A(x)>0.

I had this discussion with my maths teacher and she was very dismissive, yet offered not explanation to contradict my assertion that if I took the limit of f(x) as x->infinity and discovered that f(x) tends to A(x) and then took the limit of SQRT[f(x)] as x->infinity, I will find that SQRT[f(x)] tends to SQRT[A(x)].

My assertion may well be flawed (and to be honest I would be a tad surprised if it were not) and I would very much appreciate a detailed explanation.

 

[kony]

did you mean f(x), where f(x)>0 for all values of x for which the curve runs along the asymptote?
i mean, you can't even take the sqrt of f(x)<0.

 

[ssglain]

kony:

Provided, of course, that... f(x)>0 & A(x)>0.

 

[kony]

ssglain:

as x->infinity f(x)>0 & A(x)>0.

 

[ssglain]

Thanks a lot, Iruka. Your contribution is definitely worth a lot more than 10 cents. Hehe.