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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
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Originally posted by AThousandYoungWell, I could call it Y(a,x).
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.
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 AThousandYoung"n" is a parameter i'm going to vary for the application, but it is to be treated as a constant for this particular question.
Is n a constant or a variable?
It represents the zero of Y(x) once "a" is fixed. Y(n) = 0 and y(n) = 0
The linear equation that it transforms into as a→1 is: y(x) = b - b/n*x
Just trying to figure out the math procedure to get there, or get an idea on how these problems are typically handled.
Originally posted by AThousandYoungI know this isn't allowed, but if we take the derivative of 1 - a^(x-n) and let a=1:
Lim Y(a) as a=>1
Y(a) = (1-a^x)/(a^n-1)+1
= (1-a^x+a^n-1)/(a^n-1)
= (a^n-a^x)/(a^n-1)
Perhaps L'Hopital's Rule can be used on the above. Take the derivative of top and bottom as many times as needed?
If we take the nth derivative, we get
(1-a^(x-n))/1
lim as a=>1 of 1-a^(x-n) is...0?
1 - a^(x-n) => -(x-n)*a^(x-n-1) => -x+n => -1/n*x + 1 which is the functional limit I am after. A very nice and unfortunate coincidence I suppose.
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Originally posted by AThousandYoungHow did you get to that final form by taking n derivatives?
Lim Y(a) as a=>1
Y(a) = (1-a^x)/(a^n-1)+1
= (1-a^x+a^n-1)/(a^n-1)
= (a^n-a^x)/(a^n-1)
Perhaps L'Hopital's Rule can be used on the above. Take the derivative of top and bottom as many times as needed?
If we take the nth derivative, we get
(1-a^(x-n))/1
lim as a=>1 of 1-a^(x-n) is...0?
f(x)/g(x) = (a^n - a^x)/(a^n - 1)
f'(x)/g'(x) = (n*a^(n-1) - x*a^(x-1))/(n*a^(n-1))
f"(x)/g"(x) = (n*(n-1)*a^(n-2) - x*(x-1)*a^(x-2))/(n*(n-1)*a^(n-2))
.
.
.
fⁿ(x)/gⁿ(x) = (n! - x*(x-1)* ... *(x-n+1)*a^(x-n))/(n!)
fⁿ(x)/gⁿ(x) = 1 - (x*(x-1)* ... *(x-n+1)*a^(x-n))/n!
How does the rest simplify?
Originally posted by joe shmoSorry for the notation confusion, that should be:
How did you get to that final form by taking n derivatives?
f(x)/g(x) = (a^n - a^x)/(a^n - 1)
f'(x)/g'(x) = (n*a^(n-1) - x*a^(x-1))/(n*a^(n-1))
f"(x)/g"(x) = (n*(n-1)*a^(n-2) - x*(x-1)*a^(x-2))/(n*(n-1)*a^(n-2))
.
.
.
fⁿ(x)/gⁿ(x) = (n! - x*(x-1)* ... *(x-n+1)*a^(x-n))/(n!)
fⁿ(x)/gⁿ(x) = 1 - (x*(x-1)* ... *(x-n+1)*a^(x-n))/n!
How does the rest simplify?
f(a)/g(a) = (a^n - a^x)/(a^n - 1)
f'(a)/g'(a) = (n*a^(n-1) - x*a^(x-1))/(n*a^(n-1))
f"(x)/g"(x) = (n*(n-1)*a^(n-2) - x*(x-1)*a^(x-2))/(n*(n-1)*a^(n-2))
.
.
.
fⁿ(a)/gⁿ(a) = (n! - x*(x-1)* ... *(x-n+1)*a^(x-n))/(n!)
fⁿ(a)/gⁿ(a) = 1 - (x*(x-1)* ... *(x-n+1)*a^(x-n))/n!
Originally posted by joe shmoI might have made some mistakes. I need to think about it.
How did you get to that final form by taking n derivatives?
f(x)/g(x) = (a^n - a^x)/(a^n - 1)
f'(x)/g'(x) = (n*a^(n-1) - x*a^(x-1))/(n*a^(n-1))
f"(x)/g"(x) = (n*(n-1)*a^(n-2) - x*(x-1)*a^(x-2))/(n*(n-1)*a^(n-2))
.
.
.
fⁿ(x)/gⁿ(x) = (n! - x*(x-1)* ... *(x-n+1)*a^(x-n))/(n!)
fⁿ(x)/gⁿ(x) = 1 - (x*(x-1)* ... *(x-n+1)*a^(x-n))/n!
How does the rest simplify?
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Originally posted by joe shmoFor 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:
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
lim_{a → 1} (1 - a^x)/(a^n - 1) = lim_{a → 1} [-x*a^(x-1)]/[n*a^(n-1)] = -x/n.
So
lim_{a → 1}Y(a,x) = -x/n+1
Originally posted by SoothfastAha...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!
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:
lim_{a → 1} (1 - a^x)/(a^n - 1) = lim_{a → 1} [-x*a^(x-1)]/[n*a^(n-1)] = -x/n.
So
lim_{a → 1}Y(a,x) = -x/n+1