Let all variables represent non-negative integers. Then, given the identity: \(T*a^{x}=(T*a)^{\left({{\frac{\frac{x*\ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1}\right)\) can we let \(T=1\) ? Don.
The way you've written it, no - because you'd have a division by zero. However, multiplying the numerator and the denominator in the exponent by ln(T) puts the expression into a form where you could let T = 1 (although it's not quite clear why you'd want to do that when you already have the LHS...)
I agree. We can't let \(T=1\) Does this mean that multiplication by one automatically results in division by zero? Don.
Think of the conceptually much simpler case: f(x) = x/x A computer wouldn't accept 0 as an input to this function, even though it clearly has a sensibly defined value of 1 everywhere.
But we all just agreed that multiplying \(a^{x}\) by \(T=1\) results in division by zero in the right hand side of the identity! Thus, according to this identity, multiplication by 1 does indeed result in division by zero! Don.
f(x)= x/x at x=0 results in the indeterminate form 0/0. Indeterminate forms are allowable. Letting T=1 in the identity being discussed results in division by zero. Division by zero is not allowable. Don.
The equation is an identity. If we can't let \(T=1\) on the right, then we can't let \(T=1\) on the left either. This is true even if we somehow "transported" the term on the right all the way to the other side of the universe where no one will ever see it again! Don.
Don: Here's another identity: \(\frac{a^2 - b^2}{a-b} = a + b\) Now, suppose a = b = 7, say. Is the left-hand side of the expression valid?
Of course it is. At a=b=7, it results in the indeterminate form 0/0 which can easily be determined to be 14. Your example has absolutely nothing to do with the identity in question which does not result in an indeterminate form but a division by zero. Don.
L'Hopital's rule also applies to the indeterminate form \(\infty / \infty\), which is what the identity in your opening post has. So, I think you've proved that my example has everything to do with your one.
No, we are dealing with non-negative integers. Division by zero does not result in infinity. Division by zero is simply disallowed. Don.
Second thoughts: Ok. If division by zero is disallowed, we come back to my earlier statement in a modified form. The right-hand side of your identity is disallowed when T=1. Are we done?
Why? So, for x=0... f(x) = x/x is "allowable", but f(x) = (1/x)/(1/x) is not? Even though they're identical?
