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

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

A: 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.
A: 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, 
