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Square roots & oblique asymptotes (1 Viewer)

ssglain

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445
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lost in a Calabi-Yau
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2007
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

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Feb 10, 2006
Messages
322
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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.
 

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