Find the derivative of y = ln x by using the definition of derivative as a limit.
Answer:
We know that $e = \lim_{n \to \infty } \left(1 + \frac{1}{n}\right)^n $
So, $ 1 = ln e = \lim_{n \to \infty } \ln \left(1 + \frac{1}{n}\right)^n $
$= \lim_{n \to \infty } n \cdot \ln \left(1 + \frac{1}{n}\right)^n $
$ \frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} $
$= \lim_{h \to 0} \frac{\ln(x+h)-\ln(x)}{h} $
$= \lim_{h \to 0} \frac{\ln \frac{x+h}{x}} {h} $
$= \lim_{h \to 0} \frac {\ln \left( 1+ \frac{h}{x}\right)} {h} $
$= \lim_{h \to 0} \frac{1}{h} \ln \left( 1+ \frac{h}{x}\right) $
If we put $ m = \frac{x}{h} $, then $h = \frac{x}{m} $ and
$\frac{dy}{dx} = \lim_{m \to \infty} \frac{m}{x} \ln \left( 1+ \frac{1}{m}\right) $
$= \frac{1}{x} \cdot \lim_{m \to \infty} m \cdot \ln \left( 1+ \frac{1}{m}\right) $
$= \frac{1}{x} \cdot 1$
$= \frac{1}{x}$
