- 09 Jan '16 15:36 / 1 editWhat 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 - 09 Jan '16 20:45 / 1 edit

Well, I could call it Y(a,x).*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.**

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? - 10 Jan '16 01:45

"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.*Originally posted by AThousandYoung***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. - 10 Jan '16 06:45 / 1 editLim 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? - 10 Jan '16 16:33

I know this isn't allowed, but if we take the derivative of 1 - a^(x-n) and let a=1:*Originally posted by AThousandYoung***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. - 10 Jan '16 22:11 / 1 edit

How did you get to that final form by taking n derivatives?*Originally posted by AThousandYoung***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? - 11 Jan '16 00:33

Sorry for the notation confusion, that should be:*Originally posted by joe shmo***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! - 14 Jan '16 21:09

I might have made some mistakes. I need to think about it.*Originally posted by joe shmo***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? - 19 Jan '16 02:01 / 1 edit

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 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

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 - 21 Jan '16 00:17

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!*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:**

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