<s>[INST] Limit of $\frac{x}{x^2-1}$ Question

$$\lim_{x \rightarrow 1-}\frac{x}{x^2-1}$$

My attempt
$$\lim_{x \rightarrow 1-}\frac{x}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{x^2(\frac{1}{x^2})}{x^2(1-\frac{1}{x^2})}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{\frac{1}{x^2}}{1-\frac{1}{x^2}}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{\frac{1}{x^2}}{\frac{x^2-1}{x^2}}=\lim_{x \rightarrow 1-}\frac{x}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{x^2(\frac{1}{x^2})}{x^2(1-\frac{1}{x^2})}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{\frac{1}{x^2}}{1-\frac{1}{x^2}}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2}\cdot\frac{x^2}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}$$
As you can see I'm going in circles. Can anyone give me a hint on how to start on this problem?
 [/INST] Hint: $x^2 - 1 = (x - 1)(x + 1)$.
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