Limit definition of logarithm

138 Views Asked by At

In the book "e: The Story of a Number" (Appendix 4), the author try to show the inverse relation between $\lim_{h \to 0}\frac{b^h - 1}{h} = 1$ and $\lim_{h \to 0}\left(1 + h\right)^{1/h} = b$ in order to give an idea of the natural logarithm.

Here is how he proceeds textbook copied from the book:

Our goal is to determine the value of b for which $\lim_{h \to 0}\frac{b^h - 1}{h} = 1$. We start with the expression $\frac{b^h - 1}{h}$ for finite $h$ and set it equal to 1:

$\frac{b^h - 1}{h} = 1$. (1)

Certainly, if this expression is identically equal to 1, then also $\lim_{h \to 0}\frac{b^h - 1}{h} = 1$. We now solve equation (1) for $b$. We do this in two steps. In the first step we get

$b^h = 1 + h$

and in the second,

$b = \sqrt[^h]{\left(1 + h\right)} = \left(1 + h\right)^{1/h}$ (2)

where we replaced the radical sign with a fractional exponent. Now equation (1) expresses $b$ as an implicit function of $h$; since equations (1) and (2) are equivalent, letting $h \to 0$ will give us the equivalent expressions

$\lim_{h \to 0}\frac{b^h - 1}{h} = 1$ and $b = \left(1 + h\right)^{1/h}$.

The last limit is the number e. Thus, to make the expression $\lim_{h \to 0}\left(1 + h\right)^{1/h}$ equal to 1, $b$ must be chosen as $e = 2.71828...$

Are the steps mentioned in the book correct in regards to the laws of limits? Especially when he wrote $b^h = 1 + h \implies \left(1 + h\right)^{1/h}$ (notice how he removed the limit prefix) then he put back the limit prefix and wrote $\lim_{h \to 0}\frac{b^h - 1}{h} = 1$.

Also, he said equation (1) is an implicit function of $h$. I am not sure of what he meant, in regards to the limit definition he mentioned earlier.