# Limit of a function without using L'Hôpital Rule

Prove that

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

without using L'Hôpital Rule.

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

## 7 Answers

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.

• Good idea, but there is some whiff of circularity. – Martín-Blas Pérez Pinilla Apr 10 '14 at 14:07
• Not really, only a whiff of ambiguity since we don't know how the OP defines log. – Mikhail Katz Apr 10 '14 at 14:10

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.

HINT :

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

$$\\$$

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

• I believe if he can use McLaurin series he would not bother with avoiding l'Hôspital Rule and just go straight for an equivalent – T_O Apr 10 '14 at 14:04
• @T_O I just try to give a hint in case he didn't know. – Tunk-Fey Apr 10 '14 at 14:06
• @Tunk-Fey You beat me to it. – Josh Apr 10 '14 at 14:06
• @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. – Josh Apr 10 '14 at 14:08
• Why on earth are you putting a huge "Q.E.D." after something that is a hint not a proof? – Henning Makholm Apr 10 '14 at 17:04

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. 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.

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,

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