$\lim_{x \to 1} \frac{\ln(x^3)}{x^2-1}$ My first idea was to use the definition of derivative (but it's giving me the wrong answer). 
$\displaystyle \lim_{x \to 1}\frac{\ln(x^3)}{x^2-1} =\lim_{t \to 1}\frac{\ln(t^{3/2})}{t-1} = \frac{3}{2}\lim_{t\to 0}\frac{\ln(t+1)}{t} =\frac{3}{2}\lim_{t\to 0}\frac{\ln(t+1)-\ln(1)}{t} = \frac{d}{dx}\ln(x)\bigg|_{x=1} = 0.$
But using L'hopital's rule I get $\displaystyle \lim_{x \to 1}\frac{\ln(x^3)}{x^2-1} =\frac{3\frac{d}{dx}\ln(x)}{\frac{d}{dx}(x^2-1)} = \frac{3}{2}\lim_{x \to 1}\frac{1}{x^2} = \frac{3}{2}.$
So where have I messed up in the first approach?

Your error is that
$$\frac{d}{dx} \ln x \big|_{x = 1} = \frac 1 x \big|_{x = 1} = 1$$

Here's a slightly different way that doesn't involve making a substitution before applying the definition of the derivative:
\begin{align*}
\lim_{x \to 1} \frac{\ln(x^3)}{x^2 - 1} &= \lim_{x \to 1} \frac{3}{x + 1} \frac{\ln x}{x - 1} \\
&= \frac{3}{1 + 1} \frac{d}{dx} \ln x \big|_{x = 1} \\
&= \frac 3 2
\end{align*}
as desired. 
