Hello phish

Welcome to Math Help Forum! Quote:

Originally Posted by

**phish**

Using the laws of logarithms in the numerator:

$\displaystyle \lim_{h \to 0}\frac{\log_a(x+h) -\log_a(x)}{h}=\lim_{h \to 0}\frac{\log_a\left(\dfrac{x+h}{x}\right)}{h}$ $\displaystyle =\lim_{h \to 0}\frac{\log_a\left(1+\dfrac{h}{x}\right)}{h}$

Now when we substitute $\displaystyle u = \frac{h}{x}$, note that, in the context of this limit, it is $\displaystyle h$ that is the variable and $\displaystyle x$ a constant. So as $\displaystyle h \to 0, u \to 0$; and, of course, $\displaystyle h = ux$. So we get:$\displaystyle =\lim_{u \to 0}\frac{\log_a(1+u)}{ux}$

and, taking out the constant $\displaystyle x$ from the limit gives us the desired result:$\displaystyle =\frac{1}{x}\Big(\lim_{u \to 0}\frac{1}{u}\log_a(1+u)\Big)$

Grandad