Step by step solution to "How can I deduce that without Taylor series..."

Teddy Dillard

Answered question

2022-01-03

How can I deduce that without Taylor series or LHospitals rule?

lim_{x \to 0} \frac{\ln (1 + x)}{x} = 1

Answer & Explanation

ol3i4c5s4hr

Beginner2022-01-04Added 48 answers

If the limit exists, say

L

, then you can state that:

L = lim_{x \to 0} \frac{\ln (1 + x)}{x}

∴ e^{L} = e^{lim_{x \to 0} \frac{\ln (1 + x)}{x}}

= lim_{x \to 0} e^{\frac{\ln (1 + x)}{x}}

= lim_{x \to 0} {(e^{\ln (1 + x)})}^{\frac{1}{x}}

= lim_{x \to 0} {(1 + x)}^{\frac{1}{x}}

= e

∴ L = 1

Rita Miller

Beginner2022-01-05Added 28 answers

Note

x \geq \log (1 + x) \geq \frac{x}{1 + x}

for all

x > - 1

. Since

\frac{x}{x} \to 1

and

\frac{\frac{x}{1 + x}}{x} \to 1

x \to 0

. So the limit is

1

Vasquez

Expert2022-01-09Added 669 answers

Change the variable and use the continuity of $\ln (x)$
$\begin{matrix} lim_{x \to 0} \frac{\ln (1 + x)}{x} \\ = lim_{x \to 0} \frac{\ln (1 + \frac{1}{n})}{\frac{1}{n}} \\ = lim_{x \to 0} n \cdot \ln (1 + \frac{1}{n}) \\ = lim_{x \to 0} (\ln (1 + \frac{1}{n})^{n}) \\ \ln (lim_{x \to 0} (1 + \frac{1}{n})^{n}) \\ = \ln (e) \\ = 1 \end{matrix}$

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