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Prove that

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

without using L'Hôpital Rule.

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    $\begingroup$ That will be hard to do. $\endgroup$ – David Mitra Apr 10 '14 at 12:55
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    $\begingroup$ @DavidMitra especially because it's wrong. The limit is $0$. $ln(x) = o(x)$ $\endgroup$ – T_O Apr 10 '14 at 13:03
  • $\begingroup$ I can help you if the question is proving $$\lim_{x\to0} \frac{\ln(x+1)}{x} = 1.$$ $\endgroup$ – Tunk-Fey Apr 10 '14 at 13:11
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    $\begingroup$ Are you sure you know how to use l'hôspital rule ? $1/x$ goes to 0 when $x$ goes to $+\infty$ $\endgroup$ – T_O Apr 10 '14 at 13:40
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    $\begingroup$ Which definition of the logarithm are you working with? $\endgroup$ – Daniel Fischer Apr 10 '14 at 13:56
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This is just the derivative of $y=lnx$ at $x=1$ by using the classic definition of the derivative. And so the answer is $1$. In this way, L'Hospital is avoided.

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  • $\begingroup$ Good idea, but there is some whiff of circularity. $\endgroup$ – Martín-Blas Pérez Pinilla Apr 10 '14 at 14:07
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    $\begingroup$ Not really, only a whiff of ambiguity since we don't know how the OP defines log. $\endgroup$ – Mikhail Katz Apr 10 '14 at 14:10
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If you define $\log(1+x)$ as the integral of $\frac{1}{1+x}$ and you don't want to use the fundamental theorem of calculus either, you could bound $\frac{1}{1+x}$ between $1-\epsilon$ and $1+\epsilon$ for sufficiently small $x$, and use this to get the required limit.

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HINT :

If that limit you are asking, then you can use Maclaurin series for $\ln(1+x)$ to prove $\lim\limits_{x\to0} \frac{\ln(1+x)}{x} = 1.$

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$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$

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    $\begingroup$ I believe if he can use McLaurin series he would not bother with avoiding l'Hôspital Rule and just go straight for an equivalent $\endgroup$ – T_O Apr 10 '14 at 14:04
  • $\begingroup$ @T_O I just try to give a hint in case he didn't know. $\endgroup$ – Tunk-Fey Apr 10 '14 at 14:06
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    $\begingroup$ @Tunk-Fey You beat me to it. $\endgroup$ – Josh Apr 10 '14 at 14:06
  • $\begingroup$ @T_O I don't necessarily think he wouldn't know about Taylor series. Some calc books don't mention l'Hospital until doing sequences and series. $\endgroup$ – Josh Apr 10 '14 at 14:08
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    $\begingroup$ Why on earth are you putting a huge "Q.E.D." after something that is a hint not a proof? $\endgroup$ – Henning Makholm Apr 10 '14 at 17:04
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Integrate $\frac{1}{1+y}$ between 0 and x. Look at the picture below to get an upper and lower bound for this integral. Divide through by x and take the limit.

enter image description here

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This limit is equivalent to $$ \lim_{x \to 0} \frac{e^x-1}{x}=1, $$ which is easily handled with the definition $$ e^x = \sum_{k=0}^\infty \frac{x^k}{k!}. $$ As noticed above, it all boils down to your definition of the exponential or the logarithmic function.

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For any $x>0$ We have $$\frac{x}{x+1} =\int_{0}^x\frac{dt}{x+1} \le \int_{0}^x\frac{dt}{t+1} =\color{red}{\ln(x+1 )}=\int_{0}^x\frac{dt}{t+1} \le \int_{0}^x\frac{dt}{1} = x $$

that is for $x>0$

$$\frac{x}{x+1} \le \ln(x+1 ) \le x \implies\color{red}{\frac{1}{x+1}\le \frac{\ln(x+1 )}{x} \le 1}$$ $$\color{red}{1=\lim_{x\to 0}\frac{1}{x+1}\le \lim_{x\to 0}\frac{\ln(x+1 )}{x} \le 1}$$

that $$\color{blue}{ \lim_{x\to 0}\frac{\ln(x+1 )}{x} = 1}$$

By Squeeze Theorem, we have,

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What about this: write our the Taylor series at 0 for the numerator, cancel out terms, take the limit.

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