Given the mother rational function, $f(x)=\frac{1}{x}$, prove algebraically that $f(x)$ can never cross its horizontal asymptote.
My approach:
$f(x)=\frac{1}{x}$, $x\longrightarrow\pm\infty$, $f(x)\longrightarrow0$
$0=\frac{1}{x}$
$0=1\longrightarrow$which is not true, meaning it cannot pass it?
I am not really sure if my approach is correct, but it will be interesting to see what others think about this.
On
As I mentioned in my comment,
But if it is an asymptote, it means the function tends to it, but never reaches/crosses it. I believe your approach would be correct if you stopped after the limit you calculated. (only the first line of your approach would suffice).
I believe only calculating the limit already demonstrates:
$$\lim_{x\to\pm\infty}\frac{1}{x} = 0$$
Which means the function never reaches the asymptote $y = 0$.
This you had already done, I don't think any follow up would be necessary.
In my opinion, your proof is clear enough to be understood by the average reader. However, it is very terse and depending on what you are writing the proof for, you may need to be more clear. Here is a more elaborate formulation of your proof:
Claim: Given a function $f$ such that $f(x) = \frac{1}{x}$, prove that the curve of $f$ does not intersect its horizontal asymptote.
Proof: The function $g$ of the horizontal asymptote of the curve of $f$ is given by: $$ g(x) = \lim_{x\to\pm\infty}f(x) = \lim_{x\to\pm\infty}\frac{1}{x} = 0 $$
To demonstrate that the curve of $f$ does not intersect its horizontal asymptote, we must show that $f(x) \neq g(x) \space \forall x \in \mathbb{R}$. $$ f(x) = g(x) \implies \frac{1}{x} = 0 \implies 1 = 0x = 0 $$ Since $1 \neq 0$, we have proven by contradiction that $f(x) \neq g(x)$, and hence that the curve of $f$ does not intersect its horizontal asymptote.
Q.E.D.