Any mathfags know how to prove that a function's graph has no vertical asymptotes?

Any mathfags know how to prove that a function's graph has no vertical asymptotes?
Ex: f:R->R, f(x)=x^3-5x+2
Help a Sup Forumsro out

ask yourself what a vertical asymptote is first.

Do you perhaps mean a continuous function?

f(x) = 1/x

If this is, what you mean...

Cubic polynomial, no denominator. It's pretty clear there's no vertical asymptote.

A vertical asymptote occurs when the denominator is equal to 0. You cannot divide by 0 so the function goes to + or - infinity when x = 0. If you can prove that the denominator is never 0, then you can prove that there are no vertical asymptotes.

Shit, if english was my native language I'd probably know how to explain it. Basically, what you have as a function there is a polynome and it has no vertical ones.

You have to find for which x the function isn't defined. Your example doesn't have such x.
here that x is 0. So X=0 is the asymptote. This can be proven using limits.

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Mate, I can clearly see that it has no vertical asymptotes, but I need a fucking demonstration, doesn't work like "It has none, you dumbfucks." XD
Here's another one from the same damned exercise:
f:R->R, f(x)=x^2-2^x

f'(x)=3x^2-5. Since this derivative is bounded on R and the function is continuous on R (note that f(x) is not uniformly continuous on R as it can only be uniformly continuous on a compact interval of R), f(x) is of bounded variation. This means that f(x) is monotonic, and hence has no vertical asymptotes.

Mate you might be on to something there, lemme experiment a bit on the monotonic side...

Are you looking for an Analysis proof or something like Pre-Calc?

This

Analysis

>x = 0

I meant when the denominator = 0

A simple example is something like:

f(x) = (x^2 - 1) / (x - 1)

You might think there's a vertical asymptote at x = 1 because that would make the denominator 0, but you can simplify to cancel out the denominator, thus proving there is no vertical asymptote

f(x) = (x^2 - 1) / (x - 1)
>Factor numerator
f(x) = (x + 1)(x - 1) / (x - 1)
>(x - 1) cancels out
f(x) = (x + 1) / 1 = x + 1

this

1. Find domain
2. You realize all real numbers are in domain
3. You cannot prove it has a vertical asymptote because the function is continuous. So it has none.
What the hell are you kids doing in school. Simply writing these 3 things would be enough for me at a college level. What do you need to "prove"?

Given E>0, there exists delta > 0 s.t. |x-y| < delta => |f(x)-f(y)|

Mate, high school here is like advanced engineering in USA. We're fucking slaves. But yes, you have a good point, find the domain and prove it contains R...

I'm a mathfag. I'm assuming you're sticking to two dimensions? I.e. f:R->R and not f:R^3->R or some shit?

Strictly speaking, a vertical asymptote occurs when the limit of your function for x->a approaches infinity, with a not being +/- infinity.

If you want to take the strict math approach, you basically have to calculate this limit for any 'a'. Since you don't have a concrete value for 'a', you're either left making use of using viable mathematical rules for limits, the rule of l'hopital, or in the worst case, the pure definition of a limit.

Intuitively speaking, you can cut down the work by only considering x values that are obvious problem points. For example, every point where the function is continuously differentiable, it's not going to have an asymptote, so you won't need to bother with them.

It's a finite polynomial.