# How can I deduce that without Taylor series or L'Hospital's

How can I deduce that without Taylor series or L'Hospital's rule?
$$\displaystyle\lim_{{{x}\rightarrow{0}}}{\frac{{{\ln{{\left({1}+{x}\right)}}}}}{{{x}}}}={1}$$

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ol3i4c5s4hr
If the limit exists, say $$\displaystyle{L}$$, then you can state that:
$$\displaystyle{L}=\lim_{{{x}\rightarrow{0}}}{\frac{{{\ln{{\left({1}+{x}\right)}}}}}{{{x}}}}$$
$$\displaystyle\therefore{e}^{{L}}={e}^{{\lim_{{{x}\rightarrow{0}}}{\frac{{{\ln{{\left({1}+{x}\right)}}}}}{{{x}}}}}}$$
$$\displaystyle=\lim_{{{x}\rightarrow{0}}}{e}^{{{\frac{{{\ln{{\left({1}+{x}\right)}}}}}{{{x}}}}}}$$
$$\displaystyle=\lim_{{{x}\rightarrow{0}}}{\left({e}^{{{\ln{{\left({1}+{x}\right)}}}}}\right)}^{{\frac{{{1}}}{{{x}}}}}$$
$$\displaystyle=\lim_{{{x}\rightarrow{0}}}{\left({1}+{x}\right)}^{{\frac{{{1}}}{{{x}}}}}$$
$$\displaystyle={e}$$
$$\displaystyle\therefore{L}={1}$$
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Rita Miller
Note $$\displaystyle{x}\geq{\log{{\left({1}+{x}\right)}}}\geq{\frac{{{x}}}{{{1}+{x}}}}$$ for all $$\displaystyle{x}{>}−{1}$$. Since $$\displaystyle{\frac{{{x}}}{{{x}}}}\rightarrow{1}$$ and $$\displaystyle{\frac{{{\frac{{{x}}}{{{1}+{x}}}}}}{{{x}}}}\rightarrow{1}$$ as $$\displaystyle{x}\rightarrow{0}$$. So the limit is $$\displaystyle{1}$$.
Vasquez

Change the variable and use the continuity of $$\ln(x)$$
$$\begin{array}{}\lim_{x\rightarrow0}\frac{\ln(1+x)}{x}\\=\lim_{x\rightarrow0}\frac{\ln(1+\frac{1}{n})}{\frac{1}{n}}\\=\lim_{x\rightarrow0}n\cdot\ln(1+\frac{1}{n})\\=\lim_{x\rightarrow0}(\ln(1+\frac{1}{n})^n) \\\ln(\lim_{x\rightarrow0}(1+\frac{1}{n})^n)\\=\ln(e)\\=1 \end{array}$$