Teddy Dillard

2022-01-03

How can I deduce that without Taylor series or LHospitals rule?
$\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(1+x\right)}{x}=1$

ol3i4c5s4hr

If the limit exists, say $L$, then you can state that:
$L=\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(1+x\right)}{x}$
$\therefore {e}^{L}={e}^{\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(1+x\right)}{x}}$
$=\underset{x\to 0}{lim}{e}^{\frac{\mathrm{ln}\left(1+x\right)}{x}}$
$=\underset{x\to 0}{lim}{\left({e}^{\mathrm{ln}\left(1+x\right)}\right)}^{\frac{1}{x}}$
$=\underset{x\to 0}{lim}{\left(1+x\right)}^{\frac{1}{x}}$
$=e$
$\therefore L=1$

Rita Miller

Note $x\ge \mathrm{log}\left(1+x\right)\ge \frac{x}{1+x}$ for all $x>-1$. Since $\frac{x}{x}\to 1$ and $\frac{\frac{x}{1+x}}{x}\to 1$ as $x\to 0$. So the limit is $1$.

Vasquez

Change the variable and use the continuity of $\mathrm{ln}\left(x\right)$
$\begin{array}{}\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(1+x\right)}{x}\\ =\underset{x\to 0}{lim}\frac{\mathrm{ln}\left(1+\frac{1}{n}\right)}{\frac{1}{n}}\\ =\underset{x\to 0}{lim}n\cdot \mathrm{ln}\left(1+\frac{1}{n}\right)\\ =\underset{x\to 0}{lim}\left(\mathrm{ln}\left(1+\frac{1}{n}{\right)}^{n}\right)\\ \mathrm{ln}\left(\underset{x\to 0}{lim}\left(1+\frac{1}{n}{\right)}^{n}\right)\\ =\mathrm{ln}\left(e\right)\\ =1\end{array}$

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