What is the value of $\lim\limits_{h \to 0}\dfrac{\text{ln}(1+h)}{h}$ (with answer)

$\lim\limits_{h \to 0}\dfrac{\text{ln}(1+h)}{h}=1$ since:

(1) $\ln(a) - \ln(b) = \ln\left(\dfrac{a}{b}\right)$ since:

• Let $y = \ln(a) - \ln(b)$
• $e^y = e^{\ln(a) - \ln(b)} = \dfrac{e^{\ln(a)}}{e^{\ln(b)}} = \dfrac{a}{b}$
• $y = \ln\left(\dfrac{a}{b}\right)$

(2) $\lim\limits_{h \to 0}\dfrac{\ln(x+h) - \ln(x)}{h} = \lim\limits_{h \to 0}\dfrac{\ln\left(1 + \frac{h}{x}\right) }{h}$

(3)  Let $t = \dfrac{h}{x}$ so that $h = xt$ and:

$$\lim\limits_{h \to 0}\frac{\ln\left(1 + \frac{h}{x}\right) }{h} = \lim\limits_{t \to 0}\frac{\ln\left(1 + t\right) }{xt} = \lim\limits_{t \to 0}\left(\frac{1}{xt}\right)\ln(1+t)$$

(4)  $n\ln(a) = \ln\left(a^n\right)$ since:

• Let $y = n\ln(a)$
• $e^y = e^{n\ln(a)} = \left(e^{\ln(a)}\right)^n = a^n$
• $y = \ln\left(a^n\right)$

(5)  $\lim\limits_{t \to 0}\left(\frac{1}{xt}\right)\ln(1+t) = \lim\limits_{t \to 0}\ln\left(\left[1+t\right]^{\frac{1}{xt}}\right) = \lim\limits_{t \to 0}\ln\left(\left[\left(1+t\right)^{\frac{1}{t}}\right]^{\frac{1}{x}}\right) = \left(\dfrac{1}{x}\right)\lim\limits_{t \to 0}\ln\left(\left[1+t\right]^{\frac{1}{t}}\right)$

(6)  By definition, $e = \lim\limits_{n \to \infty}\left(1 + \dfrac{1}{n}\right)^n$ so that if $u=\dfrac{1}{t}$, then:

$$ \left(\frac{1}{x}\right)\lim\limits_{t \to 0}\ln\left(\left[1+t\right]^{\frac{1}{t}}\right) =\left(\frac{1}{x}\right)\lim\limits_{u \to \infty}\ln\left(\left[1+\frac{1}{u}\right]^{u}\right) =\left(\frac{1}{x}\right)\ln(e) = \frac{1}{x}$$

(7)  Since $\lim\limits_{h \to 0}\dfrac{\ln(x+h) - \ln(x)}{h} = \dfrac{1}{x}$, it follows that:

$$\lim\limits_{h \to 0}\frac{\ln(1+h)}{h} = \lim\limits_{h \to 0}\frac{\ln(1+h) - \ln(1)}{h} = \frac{1}{1}=1$$