1. ## derivative question

(lim x -> 0) ln(1+x)/x=1

Hi, I'm stuck on this question. The book wants me to prove it, using the definition of a derivative.

2. Originally Posted by cityismine
(lim x -> 0) ln(1+x)/x=1

Hi, I'm stuck on this question. The book wants me to prove it, using the definition of a derivative.
Ok

well we first note that $\ln(1)=0$

So then knowing that we can legally subtract zero from an expression we see that

$\lim_{\varphi\to{0}}\frac{\ln(1+\varphi)}{\varphi} =\lim_{\varphi\to{0}}\frac{\ln(1+\varphi)-\ln(1+0)}{\varphi-0}$

Now seeing that this matches up pretty well with

$f'(c)=\lim_{x\to{c}}\frac{f(x)-f(c)}{x-c}$

We can see that

$\lim_{\varphi\to{0}}\frac{\ln(1+\varphi)-\ln(1+0)}{\varphi-0}=\bigg(\ln(1+\varphi)\bigg)'\bigg|_{\varphi=0}=\ bigg(\frac{1}{\varphi+1}\bigg)\bigg|_{\varphi=0}=\ frac{1}{1+0}=1$

3. Well, just in case:

\begin{aligned}
\underset{x\to 0}{\mathop{\lim }}\,\frac{\ln (1+x)}{x}&=\underset{x\to \infty }{\mathop{\lim }}\,x\ln \left( 1+\frac{1}{x} \right) \\
& =\underset{x\to \infty }{\mathop{\lim }}\,\ln \left( 1+\frac{1}{x} \right)^{x} \\
& =\ln \Bigg[ \underset{x\to \infty }{\mathop{\lim }}\,\left( 1+\frac{1}{x} \right)^{x} \Bigg] \\
& =\ln e \\
& =1. \\
\end{aligned}