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Homework Statement
If there is a quantity T comprised of two other quantities such that T=L+B, and quantity T and B are both increasing in every period at a fixed percent such that %growth T > %growth B, it will be true that %growth L > %growth T. It will also be true that as n approaches infinity, %growth L will approach %growth T. I am trying to show this.
Homework Equations
The Attempt at a Solution
T is the sum of two other quantities:
T=L+B
In each period, T grows by the fixed percent g, while B grows by the fixed percent i, and L grows by a percent such that the first equation is true:
n=[0,\infty]
T_{0}(1+g)^n=L_{0}(1+X)^n+B_{0}(1+i)^n
To take an example, if g=.1 and i=.05, then T will grow at 10% each year while B will grow at 5% each year. Thus, L must grow at a rate X%>10%. However, as n approaches infinity, X will approach .1. I want to find X as a function of the other variables and show that it approaches g as n approaches infinity:
(1+X)^n=\frac{T_{0}(1+g)^n-B_{0}(1+i)^n}{L_{0}}
n ln(1+X) = ln\left(\frac{T_{0}(1+g)^n-B_{0}(1+i)^n}{L_{0}}\right)
ln(1+X) = \left(\frac{1}{n}\right)ln\left(\frac{T_{0}(1+g)^n-B_{0}(1+i)^n}{L_{0}}\right)
e^(ln(1+X)) = e^[(\frac{1}{n})ln(\frac{T_{0}(1+g)^n-B_{0}(1+i)^n}{L_{0}})]
(1+X) = (e^[ln(\frac{T_{0}(1+g)^n-B_{0}(1+i)^n}{L_{0}})])^\frac{1}{n}
(1+X) = [\frac{T_{0}(1+g)^n-B_{0}(1+i)^n}{L_{0}}]^\frac{1}{n}
I can't get the latex to format so that the quantities I need are being displayed as exponents, but nevertheless, I end up with:
X = [\frac{T_{0}(1+g)^n-B_{0}(1+i)^n}{L_{0}}]^(\frac{1}{n})-1
Where the fraction 1/n is in the exponent. So I think I've found X, but
lim X_{X\rightarrow \infty} = [\frac{T_{0}(1+g)^(\infty)-B_{0}(1+i)^(\infty)}{L_{0}}]^(\frac{1}{\infty})-1
The 1/infinity on the right hand side of the above equation is suppose to be an exponent. Also, the numerator of the right hand side is:
To(1+g)^infinity - Bo(1+i)^infinity
I'm not sure how to find this as an indeterminate form.