Limit of $\frac{x}{x^2-1}$

Question

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?

Answer 1

$$\lim_{x\to1^-}\frac x{x^2-1}$$ $$=\lim_{x\to1^-}\frac x{(x+1)(x-1)}$$ $$=\lim_{x\to1^-}\frac x{x+1}\cdot\frac1{x-1}$$ $$=\lim_{x\to1^-}\frac x{x+1}\cdot\lim_{x\to1^-}\frac1{x-1}$$ $$=\frac12\lim_{x\to1^-}\frac1{x-1}$$ $$\to-\infty$$

Answer 2

Hint: $x^2 - 1 = (x - 1)(x + 1)$.

Answer 3

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.

Answer 4

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 $-\infty$.

Answer 5

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 $$ \lim_{x \rightarrow 1^-} \frac{x}{x^2 - 1} = \lim_{x \rightarrow 1^-} \frac{x}{x + 1}\frac{1}{x - 1} = 0.5 \lim_{x \rightarrow 1^-} \frac{1}{x - 1} = - \infty $$