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How can I calculate this limit without Taylor series and L'Hospital's rule? $$\lim _{x\to0} \frac{\ln(1+x)}x=1$$

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If the limit exists, say $L$, then you can state that: $$\begin{align} L&=\lim_{x\to0}\frac{\ln(1+x)}{x}\\ \therefore e^L&=e^{\lim_{x\to0}\frac{\ln(1+x)}{x}}\\ &=\lim_{x\to0}e^{\frac{\ln(1+x)}{x}}\\ &=\lim_{x\to0}(e^{\ln(1+x)})^\frac{1}{x}\\ &=\lim_{x\to0}(1+x)^\frac{1}{x}\\ &=e\\ \therefore L&=1 \end{align}$$

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

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Change the variable and use the continuity of $\ln(x)$

$$\lim_{x \to 0} {\ln(1+x) \over x} = \lim_{n \to \infty} {\ln\left(1+{1 \over n}\right) \over {1\over n} }=\lim_{n \to \infty} n\cdot{\ln\left(1+{1 \over n}\right) }= \lim_{n \to \infty}{\left( \ln\left(1+{1 \over n}\right)^n \right) } $$ $$ = \ln\left( \lim_{n \to \infty}{\left( 1+{1 \over n}\right)^n }\right) = \ln(e) = 1$$

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Hint : It is equal to the caltulation of $\lim _{ x\rightarrow 0 }{ { \left( x+1 \right) }^{ \frac { 1 }{ x } } } =e$ just substitute $$t={ \left( x+1 \right) }^{ \frac { 1 }{ x } }$$

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