# Finding the Limit of a Function

joe shmo
Posers and Puzzles 09 Jan '16 15:36
1. joe shmo
Strange Egg
09 Jan '16 15:361 edit
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
2. AThousandYoung
09 Jan '16 20:22
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.
3. joe shmo
Strange Egg
09 Jan '16 20:451 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?
4. AThousandYoung
10 Jan '16 00:13
Is n a constant or a variable?
5. joe shmo
Strange Egg
10 Jan '16 01:45
Originally posted by AThousandYoung
Is n a constant or a variable?
"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.

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.
6. AThousandYoung
10 Jan '16 06:451 edit
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?
7. joe shmo
Strange Egg
10 Jan '16 16:33
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?
I know this isn't allowed, but if we take the derivative of 1 - a^(x-n) and let a=1:

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.
8. joe shmo
Strange Egg
10 Jan '16 22:111 edit
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?
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?
9. joe shmo
Strange Egg
11 Jan '16 00:33
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?
Sorry for the notation confusion, that should be:

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!
10. AThousandYoung
14 Jan '16 21:09
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?
I might have made some mistakes. I need to think about it.
11. Soothfast
0,1,1,2,3,5,8,13,21,
19 Jan '16 02:011 edit
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:

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
12. joe shmo
Strange Egg
21 Jan '16 00:17
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
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!