Prove that $\lim_{x\rightarrow \infty} \frac{x^2 - 1}{x^2 + 1} = 1$ using definition of limit.

Question

Ok, so if I have to use definition, then I should prove something like this:

$(\forall \epsilon >0)(\exists k>0)(\forall x \in X)$

then if $ x>k$ then $|f(x) - L| <\epsilon$

$L$ is the limit of the function.

Now, let's say that there is $x>k$, since this must work for any $\epsilon$ then, let's say that $k=\frac{2}{\epsilon}$

so, $ \left|\frac{x^2 - 1}{x^2 + 1} - 1\right| = \left|\frac{-2}{x^2 + 1}\right|=\frac{2}{x^2+1}<\frac{2}{x^2}<\frac{2}{x}<\frac{2}{k}=\epsilon$

Now, I'd like to know if this is correct, since I rarely do limits by definition, so I am not quite sure about this.

Answer 1

The mathematical details are correct. I would perhaps work a bit with the wording. This is how I would write it, were I given that problem on a test or as homework:

Given an $\epsilon > 0$, choose $k$ such that $k > \frac2{\epsilon}$ and $k>1$. Then we have for any $x > k$:$$\left|\frac{x^2 - 1}{x^2 + 1} - 1\right|= \cdots <\frac 2k < \epsilon$$Since $\epsilon$ was arbitrary, this proves that the function does converge to $1$.

Note that $\frac{2}{x^2} < \frac 2x$ is not true if $x \leq 1$, so you need to be specific about that too.

Answer 2

here is short cut method. simply factor out $x^2$

$\lim_{x\to \infty}\frac{x^2-1}{x^2+1}=\lim_{x\to \infty}\frac{1-1/x^2}{1+1/x^2}=\frac{1}{1}=1$