Limit at infinity, no vert asymptote, why?

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SUMMARY

The discussion centers on evaluating the limit of the function \(\frac{\sqrt{4x^2+2x+10}-4}{x-1}\) as \(x\) approaches 1 and understanding the absence of a vertical asymptote. The user correctly identifies that the limit does not approach \(\pm\infty\) as \(x\) approaches 1, leading to the conclusion that there is no vertical asymptote. The solution involves recognizing that multiplying by the conjugate, \(\sqrt{4x^2+2x+10}+4\), simplifies the expression and clarifies the limit behavior.

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  • Understanding of limits in calculus
  • Familiarity with vertical and horizontal asymptotes
  • Knowledge of algebraic manipulation, specifically using conjugates
  • Graphing functions to analyze behavior near critical points
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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?
 
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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.
 
Thanks, I always seem to forget the simple things.
 

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