Limit at infinity, no vert asymptote, why?

[jrjack]

Homework Statement

Find the vertical asymptote(n) and evaluate the limit as [itex]x \rightarrow n^-, x\rightarrow n^+[/itex], or state Does Not Exist.

Homework Equations

[tex]\frac{\sqrt{4x^2+2x+10}-4}{x-1}[/tex]

The Attempt at a Solution

I have taken the limits at [itex]\pm\infty=2,-2[/itex] and understand those are my horizontal asymptotes.

I have also finished the problem and got the correct answer (DNE), but I cannot mathematically understand why it does not have a vertical asymptote, I figured this out by graphing.

Based on similar problems, I (wrongly) assumed setting the denominator equal to 0 gives the Vert Asymptote. This was the case in the 3 problems before this one, but that was after factoring and reducing the equations. After trying to find the limit of this equation as [itex]x\rightarrow1[/itex]...I gave up, and graphed it.

I don't feel this is the correct approach. What is a better approach? Can I find that this problem has no vertical asymptote without graphing?

...After typing this I think my answer lies in the definition of a vertical asymptote, and since the limit of f(x) as x->1 was not [itex]\pm\infty[/itex], then there is no vert. asymptote.

Is that correct?

 

[Dick]

Sure, it's not a vertical asymptote because the limit as x->1 is not infinity. To work this out algebraically multiply the numerator and denominator by sqrt(4x^2+2x+10)+4 and factor. It's the usual 'conjugate' thing you should think about when you see a square root.

 

[jrjack]

Thanks, I always seem to forget the simple things.