Proving that $\lim \frac{x}{ x^2-1} = + \infty $ as $x \rightarrow 1^{+}$ Using the limit definition.

Question

Proving that $\lim \frac{x}{ x^2-1} = + \infty $ as $x \rightarrow 1^{+}$

My Attempt: I arrived at $\frac{x}{ x^2-1} > M$, then I thought of adding +1 and -1 to the numerator, but then what I will still has 2 fractions one contain x and the other contain x^2, what shall I do?

Answer 1

Hint:

$$\frac{x}{x^2-1} = \frac{x}{(x-1)(x+1)} = \frac{1}{x-1}\cdot \frac{x}{x+1}$$

also, when $x$ is close to $1$, the expression $\frac{x}{x+1}$ is bounded and close to $\frac12$ (away from $0$ is all that really matters here)

Answer 2

You may set $x= 1+ \varepsilon$ with $0 < \varepsilon < 1$.

Now write $$\frac{x}{ x^2-1} = \frac{x-1+1}{ (x+1)(x-1)}=\frac{1}{x+1}\left(1+\frac{1}{x-1}\right) \stackrel{x=1+\varepsilon}{=}\frac{1}{2+\varepsilon}\left(1+\frac{1}{\epsilon}\right)\stackrel{0 < \varepsilon < 1}{>}\frac{1}{3\varepsilon}$$

Answer 3

For fun:

Let $x >1$.

Consider $g(x)= \dfrac{x^2-1}{x}= x -1/x >0$ for $x>1$.

$\lim_{x \rightarrow 1^+}g(x) =0$. (Recall $g(x) >0$)

Hence

$\lim_{ x \rightarrow 1^+} \dfrac{1}{g(x)} = \infty.$