(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 $\displaystyle \ln(1)=0$
So then knowing that we can legally subtract zero from an expression we see that
$\displaystyle \lim_{\varphi\to{0}}\frac{\ln(1+\varphi)}{\varphi} =\lim_{\varphi\to{0}}\frac{\ln(1+\varphi)-\ln(1+0)}{\varphi-0}$
Now seeing that this matches up pretty well with
$\displaystyle f'(c)=\lim_{x\to{c}}\frac{f(x)-f(c)}{x-c}$
We can see that
$\displaystyle \lim_{\varphi\to{0}}\frac{\ln(1+\varphi)-\ln(1+0)}{\varphi-0}=\bigg(\ln(1+\varphi)\bigg)'\bigg|_{\varphi=0}=\ bigg(\frac{1}{\varphi+1}\bigg)\bigg|_{\varphi=0}=\ frac{1}{1+0}=1$
Well, just in case:
$\displaystyle \begin{aligned}
\underset{x\to 0}{\mathop{\lim }}\,\frac{\ln (1+x)}{x}&=\underset{x\to \infty }{\mathop{\lim }}\,x\ln \left( 1+\frac{1}{x} \right) \\
& =\underset{x\to \infty }{\mathop{\lim }}\,\ln \left( 1+\frac{1}{x} \right)^{x} \\
& =\ln \Bigg[ \underset{x\to \infty }{\mathop{\lim }}\,\left( 1+\frac{1}{x} \right)^{x} \Bigg] \\
& =\ln e \\
& =1. \\
\end{aligned}$