How does one show that $\frac{x^2 + 1}{x+1}$ is $O(x)$?

Question

I found this pdf solution to the problem (see bottom of page 2), but I don't understand their approach. I understand how they used algebra to rewrite the function in simpler terms, but I don't quite understand how to prove that:

$x - 1 + \frac{2}{x+1} \leq cx$ for $c = 1$ and $x > 1$

Here's what I did:

Let $x \geq 1$:

$x + 1 \geq 2$

$\frac{1}{x+1} \leq \frac{1}{2}$

$\frac{2}{x+1} \leq 1 \leq x$

So I rewrote the inequality as:

$x - 1 + \frac{2}{x+1} \leq x + 0 + x $

The right side is term-for-term greater than or equal to the left side.

$x - 1 + \frac{2}{x+1} \leq 2x$, where the witness $c = 2$

Is my approach valid? How does the pdf approach work? Any help or advice would be greatly appreciated.

Answer 1

$f(x) \in O(g(x)) \iff \|f(x)\| < M \|g(x)\|$ for $x \to \infty$ Now $$ \frac{x^2 + 1}{x(x+1)} \to 1$$ hence ... $\frac{x^2 + 1}{x+1} < 2 x$ for $x$ large enough.

Suppose that there are infinitely many $k \in \mathbb{N}$ such that $\frac{x_k^2+1}{x_k+1} \geq 2x_k$ hence $\frac{x_k^2 + 1}{x_k(x_k+1)} \geq 2$ which is a contradiction with the fact that $\frac{x_k^2 + 1}{x_k(x_k+1)} \to 1$