How do I find the limit of $\frac{x \cdot \ln(1+x)}{e^{2x}-1}$ How do I find: $$\lim_{x \to 0} \frac{x \cdot \ln(1+x)}{e^{2x}-1}$$  without using L'Hospital.
I know that $$\lim_{x \to c}\frac{f(x)}{g(x)} = \frac{L}{M}, M \neq 0$$ 
My problem is I get: $M = 0$
I've also tried to factor, but can't to get rid of the denominator.

Extra question: Geogebra gets an intersection point between the function and x-axis at $(0,0)$, shouldn't the function be discontinuous since $M = 0$?  

 A: Factor the denominator as $(e^x - 1)(e^x + 1)$ (difference of squares) and then you need to know that
$$ \lim_{x \to 0} \frac{e^x - 1}{x} = 1. $$
Which you will have seen when you first encountered the derivative of $e^x$ via first principles. This should look familiar to you:
$$ \frac{d}{dx} e^x = \lim_{h \to 0} \frac{e^{x + h} - e^h}{h} = \lim_{h \to 0} \frac{e^x(e^{h} - 1)}{h} = e^x \lim_{h \to 0} \frac{e^{h} - 1}{h}. $$
A: Following Thomas Andrew's suggestion in the comments, re-write the expression as
$$\frac{x \ln(1+x)}{e^{2x} - 1} = \frac{\ln (1+x)}{\frac{e^{2x}-1}{x}}.$$
Let's try to evaluate the limit using the division rule for limits on the second expression.  The numerator is continuous, so $\lim_{x \to 0} \ln(1+x) = \ln (1) = 0$.  For the denominator, notice that $e^{2\cdot 0} = 1$, so by the definition of the derivative as a difference quotient
$$\lim_{x\to0}\frac{e^{2x}-1}{x} = \left.\frac{d}{dx} e^{2x}\right|_{x = 0} = 2.$$
Thus,
$$\lim_{x\to 0}\frac{x \ln(1+x)}{e^{2x} - 1} = \frac{\lim_{x\to0} \ln(1+x)}{\lim_{x\to0}\frac{e^{2x}-1}{x}} = \frac{0}{2} = 0.$$

As for the extra question, it appears Geogebra is somehow computing the limit at $0$.
You are right that evaluating the function gives a $\frac{0}{0}$ indeterminate form, so strictly speaking $f(x) = \frac{x\ln(1+x)}{e^{2x} -1 }$ is undefined at $x = 0$.  But, since $\lim_{x \to 0} f(x)$ is defined, we can 'fill in' the missing point $(0,0)$ in the graph to get a graph of a continuous function, which we call the extension $\tilde f$:
$$\tilde f(x) = \begin{cases}
  \frac{x\ln(1+x)}{e^{2x} -1} & x \neq 0\\
  0 & x = 0
\end{cases}$$
Points $a$ where $f$ is undefined but $\lim_{x \to a} f(x)$ is defined are called removable discontinuities.
A: Using L-hospital
$$\lim_{x \to 0} \frac{x \cdot \ln(1+x)}{e^{2x}-1}$$
Now as !
approaches to 0 it is in $\frac{0}{0}$ so just apply 
L'Hospital's rule differentiating numerator and denominator we get
$$\lim_{x \to 0} \frac{\ln(1+x)+\frac{x}{1+x}}{e^{2x}×2}$$
Which is $0/1=0$
Without L-hospital
As strants has provided the exact solution just by defination still this also works 
As $$\frac{1}{1-x}=1+x+x^{2}+x^{3}+....$$
$$\frac{1}{1+x}=1-x+x^{2}-x^{3}+....$$
Integrating both sides 
     $$\ln(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}....$$
   Similarly $e^{2x}=1+2x+\frac{(2x)^{2}}{2!}+....$
At 
$$\lim_{x \to 0} \frac{x \cdot \ln(1+x)}{e^{2x}-1}$$
$$\lim_{x \to 0} \frac{x^{2}-\frac{x^{3}}{2}+\frac{x^{4}}{3}...}{1+2x+2x^2}$$ which is clearly $\frac{0}{1}=0$
A: $$\lim_{x\rightarrow0}\frac{x\ln(1+x)}{e^{2x}-1}=\lim_{x\rightarrow0}\left(\frac{\ln(1+x)}{x}\cdot\frac{2x}{e^{2x}-1}\cdot\frac{x}{2}\right)=1\cdot1\cdot0=0.$$
