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?
 A: $$\lim_{x\to1^-}\frac x{x^2-1}$$
$$=\lim_{x\to1^-}\frac x{(x+1)(x-1)}$$
$$=\lim_{x\to1^-}\frac x{x+1}\cdot\frac1{x-1}$$
$$=\lim_{x\to1^-}\frac x{x+1}\cdot\lim_{x\to1^-}\frac1{x-1}$$
$$=\frac12\lim_{x\to1^-}\frac1{x-1}$$
$$\to-\infty$$
A: Hint: $x^2 - 1 = (x - 1)(x + 1)$.
A: Hint:
We can do partial fraction decomposition to show that
$$\frac x{x^2-1}=\frac12\left(\frac1{x-1}+\frac1{x+1}\right)$$
Then do the limit on each individual piece.
A: hint:
factor the denominator as follows
$x^2-1=(x-1)(x+1)$.
the numerator $x$ goes to $1$.
in the denominator, $(x+1)$ goes to $2$ but $(x-1)$ tends to $0$.
so, the denominator goes to $0$.
but $x$ is less than $1$, so the denominator goes to $0$ by negatives values, let's say it tends to $0^-$.
thus the final limit is $"\frac{1}{0^-}"$
which means $-\infty$.
A: I think one can prove that if f(x0) is finite, then
$$
\lim_{x->x_0}f(x)g(x) = f(x_0) \lim_{x->x_0}g(x)
$$
One can thus avoid partial fractions by substitution
$$
  \lim_{x \rightarrow 1^-} \frac{x}{x^2 - 1} = 
  \lim_{\epsilon \rightarrow 0^+} \frac{1 - \epsilon}{(1 - \epsilon)^2 - 1} = 
  \lim_{\epsilon \rightarrow 0^+} \frac{1}{\epsilon}\frac{1 - \epsilon}{\epsilon - 2} = 
  \lim_{\epsilon \rightarrow 0^+} -\frac{1}{2\epsilon} = - \infty
$$
Or even a direct evaluation
$$
  \lim_{x \rightarrow 1^-} \frac{x}{x^2 - 1} = 
  \lim_{x \rightarrow 1^-} \frac{x}{x + 1}\frac{1}{x - 1} = 
  0.5 \lim_{x \rightarrow 1^-} \frac{1}{x - 1} =  - \infty
$$
