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

Teddy Dillard 2022-01-03 Answered
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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Expert Answer

ol3i4c5s4hr
Answered 2022-01-04 Author has 1471 answers
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
Answered 2022-01-05 Author has 233 answers
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 \(\displaystyle{1}\).
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Vasquez
Answered 2022-01-09 Author has 8850 answers

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

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