# $\lim_{x\rightarrow 0}\frac{\log(x+1)}{x}=1$

How can I prove it without using Taylor series, L'Hopital's rule, or integrals,or series expansion? I don't know how. Please help.

First I have tried to use epsilon-delta, I also tried to use sequences but I failed.

• Are you allowed to use L'Hopital's Rule? – Kavi Rama Murthy Dec 14 '19 at 5:42
• No. @KaboMurphy – New2Math Dec 14 '19 at 5:43
• What is your definition of the logarithm? – Martin R Dec 14 '19 at 6:09
• Then how do you want to prove it without using Taylor series? Note that the definition of $\exp$ as a Taylor series easily implies $\lim_{y \to 0}\frac{e^y-1}{y} = 0$, and that implies your desired limit. – Martin R Dec 14 '19 at 6:19
• I suggest that you clarify your question with the necessary information (such as how log/exp are defined). Requiring “without using Taylor series” does not make much sense if the definitions are based on Taylor series. – Also note that your question should be clear without the title. – Martin R Dec 14 '19 at 6:45

The Inequalities section of Wikipedia's "List of Logarithmic Identities" page states

$$\frac{x}{\sqrt{1+x+x^2/12}} \le \ln(1 + x) \le \frac{x}{\sqrt{1+x}} \tag{1}\label{eq1A}$$

for $$0 \le x$$, and the reverse for $$-1 \lt x \le 0$$. Dividing by $$x$$ and then having $$x \to 0^{+}$$ as shown, and with the reverse inequalities for $$x \to 0^{-}$$, gives $$1$$ on either side in both cases. Thus, by the Squeeze Theorem, you have

$$\lim_{x \to 0}\frac{\log(1+x)}{x} = 1 \tag{2}\label{eq2A}$$

• Does not help me – New2Math Dec 14 '19 at 6:10
• @New2Math This answers your question as you've written it. Since it doesn't help you, I suggest you reword your question to be more explicit about what you are willing to, and not willing to, accept for an answer. Thanks. – John Omielan Dec 14 '19 at 6:14
• It is too difficult – New2Math Dec 14 '19 at 6:16

Since $$\mathop {\lim }\limits_{x \to \pm \infty } \left( {1 + \frac{1} {x}} \right)^x = e$$ you have that $$\mathop {\lim }\limits_{x \to \pm \infty } x\log \left( {1 + \frac{1} {x}} \right) = 1$$ Now, let be $$t=\frac{1}{x}$$. You have that $$\mathop {\lim }\limits_{t \to 0} \frac{1} {t}\log \left( {1 + t} \right) = 1$$

• We have not talked about this identity of e – New2Math Dec 14 '19 at 6:17

Still calculus approach, just no integrals or explicit L'Hopital:

Let $$\varphi(x)=\log(x+1)-x$$ for $$x>0$$, then $$\varphi'(x)<0$$ for $$x>0$$, so $$\varphi(x)\leq\varphi(0)=0$$, we get $$\dfrac{\log(1+x)}{x}\leq 1$$ for $$x>0$$.

On the other hand, let $$\varphi(x)=\log(x+1)-x+x^{2}/2$$ for $$0, then $$\varphi'(x)>0$$ for $$0, so $$\dfrac{\log(x+1)}{x}\geq 1-\dfrac{1}{2}x$$ for $$0.

Finally, one concludes by Squeeze Theorem.

For negative part, we exploit the trick like the following: \begin{align*} \lim_{x\rightarrow 0^{-}}\dfrac{\log(x+1)}{x}&=\lim_{y\rightarrow 0^{+}}\dfrac{\log(1-y)}{-y}\\ &=\lim_{y\rightarrow 0^{+}}\dfrac{-\log(1-y)}{y}\\ &=\lim_{y\rightarrow 0^{+}}\dfrac{\log\left(\dfrac{1}{1-y}\right)}{y}\\ &=\lim_{y\rightarrow 0^{+}}\dfrac{\log\left(1+\dfrac{y}{1-y}\right)}{y}\\ &=\lim_{y\rightarrow 0^{+}}\dfrac{\log\left(1+\dfrac{y}{1-y}\right)}{\dfrac{y}{1-y}}\dfrac{1}{1-y}\\ &=1\cdot 1\\ &=1. \end{align*}

• What is $\varphi'$ I also don't know that $e= \lim (n+1/n)^n$ – New2Math Dec 14 '19 at 6:13
• I didn't use the limit fact. And $\varphi'$ is the derivative of $\varphi$, I use the derivative criterion to show the monotonicity of the function $\varphi$. – user284331 Dec 14 '19 at 6:27
• But we haven't talked about derivatives yet – New2Math Dec 14 '19 at 6:29
• Then I really don't know how to help, sorry. – user284331 Dec 14 '19 at 6:29

Since you have stated you know that exp(x) = $$\Sigma_{n=0}^{\infty}\frac{x^n}{n!}$$

Let $$\log(1+x) = a \implies e^a-1=x$$

as $$x \to 0 \ , a\to0$$

Now, the question rephrases to $$\lim_{a \to 0}\frac{a}{e^a-1}$$ , Now using the expansion of exp(a) the limit evaluates to 1.

Hope this helps.

In the comments you appear to accept that $$\lim_{y\to0}\frac{e^y-1}y=1.$$ As $$x\to0$$, $$1+x\to1$$ and so $$\ln(1+x)\to0$$. Therefore setting $$y=\ln(1+x)$$ gives $$1=\lim_{y\to0}\frac{e^y-1}y=\lim_{x\to0}\frac{(1+x)-1}{\ln(1+x)} =\lim_{x\to0}\frac{x}{\ln(1+x)}$$ etc.

Hint For $$x >0$$ whenever when $$n \leq \frac{1}{x} you have $$\log (1+\frac{1}{n+1})^n\leq \frac{\log(x+1)}{x} \leq \log (1+\frac{1}{n})^{n+1}$$

Use the fact that

$$\lim_n (1+\frac{1}{n+1})^n= \lim_n (1+\frac{1}{n})^{n+1}=e$$

Edit With your definition of $$e$$.

Hint 1: Use your definition of $$e$$ to show that $$\lim_{x \to 0} \frac{e^x-1}{x} =e$$

Hint 2 If you meet it, apply the formula for $$(f^{-1})'(0)$$. If not, calculate $$\lim_{x\rightarrow 0}\frac{\log(x+1)}{x}$$ via the substitution $$x=e^y-1$$. Then, you get $$\lim_{x\rightarrow 0}\frac{\log(x+1)}{x}=\lim_{y\rightarrow 0}\frac{\log(e^y-1+1)}{e^y-1}$$

• The last fact is not written in the book. We have defined e as exp(1) where exp(x)= $\sum x^n/n!$ we also know that exp(x)= $\sum^n x^n/n!+ R_{n+1}$ with $|R_{n+1}|\leq 2|x^{n+1}/(n+1!)|$ with $x\leq 1+0.5n$ now help me please. I had to write down everything on mobile phone because my computer is broken – New2Math Dec 14 '19 at 6:01
• @New2Math The usual definition of $e$ is the one in my post. I will correct the answer to use your definition. – N. S. Dec 14 '19 at 18:27