- #1
henpen
- 50
- 0
Reading Cauchy's lecture on the derivative, I see he goes from this limit
[tex]\large \lim _{h \rightarrow 0} \frac{1}{ \log_A((A^h)^{\frac{1}{A^h-1}})}[/tex]
To this one [itex]A^h= 1+\beta[/itex]
[tex]\large \lim _{\beta \rightarrow 0} \frac{1}{ \log_A((1+ \beta)^{\frac{1}{\beta}})}=\frac{1}{\log_A(e)}[/tex]
I understand the intuition behind the technique, and the result. However, is this variable-change rigorous? How can we be sure that [itex] \beta =A^h-1 [/itex] tends to 0 in the same way as [itex] h[/itex] does, or do we just need to know that when [itex] h=0 \Rightarrow \beta=0[/itex], so the limit will be the same?
[tex]\large \lim _{h \rightarrow 0} \frac{1}{ \log_A((A^h)^{\frac{1}{A^h-1}})}[/tex]
To this one [itex]A^h= 1+\beta[/itex]
[tex]\large \lim _{\beta \rightarrow 0} \frac{1}{ \log_A((1+ \beta)^{\frac{1}{\beta}})}=\frac{1}{\log_A(e)}[/tex]
I understand the intuition behind the technique, and the result. However, is this variable-change rigorous? How can we be sure that [itex] \beta =A^h-1 [/itex] tends to 0 in the same way as [itex] h[/itex] does, or do we just need to know that when [itex] h=0 \Rightarrow \beta=0[/itex], so the limit will be the same?