<s>[INST] Computing $\lim_{x \to 1}\frac{\log(x)}{x^2+x-2}$ I'm trying to compute the following limit and would greatly appreciate your feedback:
$$\lim_{x \to 1}\frac{\log(x)}{x^2+x-2}$$
Given that $\log(x)$ can be replaced with $\log(x+1)$ because of the identity
$$\log(x) + \log(1) = \log(x) + 0$$
Then using the series:
$$\log(x+1) = x  + o(x^n)$$
Then dividing by $x^2+x-2$, I get:
$$\frac{x}{x^2+x-2} = \frac{x + o(x^n)}{(x-1)(x+2)}$$
when $x = 1$
$$\frac{x + o(x^n)}{(x-1)(x+2)} = \frac{1}{0} = 0$$
hence
$$\lim_{x \to 1}\frac{\log(x)}{x^2+x-2} \implies 0$$
 [/INST] Let $x = e^t$. Then $t\to 0$ as $x\to 1$. The limit then becomes
$$\lim_{x \to 1}\frac{\log(x)}{x^2+x-2} = \lim_{t\to0}\frac{t}{e^{2t}+e^t-2} = \lim_{t\to0}\color{red}{\frac{t}{e^t-1}}\frac{1}{e^t+2} = \color{red}1\cdot \frac{1}{1+2} = \frac{1}{3}$$
$\color{red}{\text{Using}}$ the well-known limit $\lim_{x\to0}\frac{e^x-1}{x}=1$.
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