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Without using L'Hopital's Rule, how does one find $\lim_{x \to 0^{+}} \frac{\ln(x+1)}{x}$?

I was hoping to find a way using basic calc I, pre-differentiation knowledge and not knowing the definition of $e$--much like you can prove $\lim_{x\to\infty} \frac{\sin x}{x} = 1$ using a geometric argument and the squeeze rule/theorem.

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Do you know how to find $\lim\limits_{h\to 0}\dfrac{f(0+h)-f(0)}{h}$ when $f(x)=\ln(x+1)$? I.e., do you know how to find the derivative of $\ln(x+1)$ at $0$? What is your definition of $\ln$? – Jonas Meyer Sep 18 '12 at 3:51

The answer depends on what properties of $\ln$ you can use. If you know that the derivative of $\ln x$ is $1/x$, then you can use $$ \lim_{x\rightarrow 0} \frac{\ln (1+x)}{x} = \lim_{x\rightarrow 0} \frac{\ln (1+x) - \ln 1}{x} = \left.\frac{ d\ln x}{dx}\right|_{x=1} = \left.\frac{1}{x}\right|_{x=1}=1. $$

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Indeed, the stated limit is precisely the definition of the derivative of $\log'(1)$. – Lubin Sep 18 '12 at 4:38

If you know that $\lim_{x\to\infty}\left(1+\frac1x\right)^x=e$, either because this is how $e$ was defined or because you’ve already proved it, you can begin by letting $L=\lim_{x\to 0^+}\frac{\ln(1+x)}x$. The exponential function is continuous, so

$$e^L=e^{\lim_{x\to 0^+}\frac{\ln(1+x)}x}=\lim_{x\to 0^+}e^{\frac{\ln(1+x)}x}=\lim_{x\to 0^+}(1+x)^{1/x}=\lim_{x\to\infty}\left(1+\frac1x\right)^x=e\;,$$

and hence $L=\ln e=1$.

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  1. For every $x\geqslant0$, $\displaystyle\ln(1+x)=\int_1^{1+x}\frac{\mathrm dt}t$.
  2. For every $t$ in $[1,1+x]$, $\displaystyle\frac1{1+x}\leqslant\frac1t\leqslant1$.

Which implies that $\displaystyle\frac{x}{1+x}\leqslant\ln(1+x)\leqslant x$, and the desired limit.

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You could use the series expansion: $$\ln(x+1)=x-x^2/2+x^3/3-x^4/4\ .\ .\ .$$

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The OP hasn't explicitly stated so, but if he is finding the limit $\lim_{x \to 0} \dfrac{\ln(x+1)}{x}$, he is likely looking for how to evaluate $f'(1)$ when $f(x) = \ln(x)$. In this situation your reason is circular. – JavaMan Sep 18 '12 at 5:54

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