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

Question

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.

Answer 1

Hint For $x >0$ whenever when $n \leq \frac{1}{x} <n+1$ 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 $$\lim_{x\rightarrow 0}\frac{\log(x+1)}{x}=\lim_{y\rightarrow 0}\frac{\log(e^y-1+1)}{e^y-1}$$

Answer 2

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 $$ \mathop {\lim }\limits_{t \to 0} \frac{1} {t}\log \left( {1 + t} \right) = 1 $$

Answer 3

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 $0<x<1$, then $\varphi'(x)>0$ for $0<x<1$, so $\dfrac{\log(x+1)}{x}\geq 1-\dfrac{1}{2}x$ for $0<x<1$.

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*}

Answer 4

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

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

Answer 5

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.

Answer 6

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.