Normally limits don't work like that. If the function is Y(x) then you would have x approach some limit, not a different variable.
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09 Jan '16 20:45>1 edit
Originally posted by AThousandYoung Normally limits don't work like that. If the function is Y(x) then you would have x approach some limit, not a different variable.
Well, I could call it Y(a,x).
The limit of Y(a,x) as a→1 = y(a→1,x) = m(b,n)*x+b
where m(b,n) is the slope of the functional limit, and is itself a function of n, and b.
I understand its a bit different from a typical limit, but it appears that a functional limit exists, and is of this form. So how does one get there?
Originally posted by joe shmo What is the proper way to find the limit of this function Y(x) as "a" → 1
Y(x) = [1/( a^n - 1)] * ( 1 - a^x ) + 1
where "n" is a constant.
Lets let the limit of Y(x) as "a" → 1 = y(x)
Graphically,I know that y(x) is linear in "x", but I cant seem to find the correct mathematical approach to get there.
Thanks
For n and x fixed (n not equal to 0), and "a" the only variable, you can use L'Hopital's Rule (differentiating with respect to a) just once to evaluate the limit:
Originally posted by Soothfast For n and x fixed (n not equal to 0), and "a" the only variable, you can use L'Hopital's Rule (differentiating with respect to a) just once to evaluate the limit:
Aha...There it is! Haven't thought about finding limits for some time, and it turns out we were on wild goose chase with the multiple derivatives. Thank You Soothfast!