ssglain
Member
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.
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.