How are these two equivalent? $$\frac{\ln(e^x+x)}{x}=\frac{e^x+1}{e^x+x}$$
I see that they did something to get rid of the natural log. I couldn't find any properties that would allow me to do this. I also think that they raised both the numerator and denominator by $e$. I have tried it and I did not get the same result. Does anyone know how the solution manual got this? I am supposed to use L'Hopital's rule to find the limit as x approaches $0$.
 A: $$\frac{\ln(e^x+x)}{x} \; \neq \;\;\frac{e^x+1}{e^x+x}$$
It's the application of l'Hospital's rule to the left hand side: take the derivative of the LHS's numerator and of the denominator.
$$\large \frac{e^x+1}{e^x+x}=\frac{\frac{e^x+1}{e^x+x}}{1} = \frac{\frac d{dx}\left({\ln(e^x+x)}\right)}{\frac d{dx}(x)}$$ 
So $$\lim_{x\to 0} \frac{\ln(e^x+x)}{x} = \lim_{x\to 0} \frac{e^x+1}{e^x+x}$$
A: The equation that you wrote is false. The correct statement is an application of l’Hospital’s rule:
$$\lim_{x\to 0}\frac{\ln(e^x+x)}x=\lim_{x\to 0}\frac{e^x+1}{e^x+x}\;.$$
Since both $\ln(e^x+x)$ and $x$ approach $0$ as $x\to 0$, you can apply l’Hospital’s rule to evaluate the limit. To do this you must differentiate the numerator and the denominator separately.
$$\frac{d}{dx}\ln(e^x+x)=\frac{\frac{d}{dx}(e^x+x)}{e^x+x}=\frac{e^x+1}{e^x+x}\;,$$
and of course $\frac{dx}{dx}=1$, so 
$$\lim_{x\to 0}\frac{\ln(e^x+x)}x=\lim_{x\to 0}\frac{\frac{e^x+1}{e^x+x}}1=\lim_{x\to 0}\frac{e^x+1}{e^x+x}\;.$$
It is not true without the limits.
