Calculating the limit as $x$ approaches $0$ using L'Hospital's rule or series expansion is straightforward, but how to evaluate the limit without either of those techniques.
How to calculate the following limit as $x$ approaches $0$:
$\dfrac{\ln(x+1)+1-e^x}{x^2}$
This question boils down to showing that the limit $$L=\lim _{x\to 0}\frac{e^x-1-x}{x^2}$$ exists. Using $e^x-1=t$ we can see that the above implies that $$L=\lim_{x\to 0}\frac{x-\log(1+x)}{x^2}$$ and adding this to the first limit we get the desired limit in question as $-2L$. You can use binomial theorem and the definition $$e^x=\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n$$ to get $L=1/2$.