$$\lim _{ x\rightarrow 9 }{ \frac { x^{ 1/2 }+x-6 }{ x^{ 3/2 }-27 } } $$

I have to state whether or not the limit requires L'Hopital's Rule to be evaluated. However, before I can do that, I must see check if it is in indeterminate form. This is where I ran into an issue. When I evaluate it at first I get:

$$\lim _{ x\rightarrow 9 }{ \frac { (9)^{ 1/2 }+(9)-6 }{ (9)^{ 3/2 }-27 } } =\frac { 3+9-6 }{ 27-27 } =\frac { 6 }{ 0 } =\infty $$

I think my reasoning is correct to assume that the one sided limit is positive infinity. However, I don't understand why the limit from the other side is negative infinity. Is this because It can either be the principal or negative square root of $9$? This is where i got confused.

  • $\begingroup$ No, it's just because $\sqrt x+ x<6$ when $x<9$ (note the function is increasing) $\endgroup$ – Adam Hughes Mar 4 '15 at 0:57
  • $\begingroup$ $\sqrt 8 + 8 < 6$? It is the denominator that is responsible for the change in sign. $\endgroup$ – BaronVT Mar 4 '15 at 1:02

It's a function with a vertical asymptote at $x = 9$.

In some cases, a function like $\dfrac{1}{x^2}$ has $$\lim_{x\to 0} \dfrac{1}{x^2} = \infty$$ (i.e. the limit is positive infinity from the left and the right). Sometimes it doesn't; $\dfrac1x$ is an obvious example.

In this case, the denominator changes sign at $x = 9$ (as you have hopefully noticed), and the numerator does not, and this is reflected in the limits.

You have done everything correctly: you have determined that the limit is not finite, and the right and left side limits are not the same (so really "the limit does not exist" or "the limit from the right is positive infinity and from the left it is negative infinity" are about the best you can do), and you did not need L'Hopital's rule at any point.

  • $\begingroup$ Just to recap: I should have noted the vertical asymptote at $x=9$ and given this, I should've realized how the function splits off; heading to positive and negative infinity. $\endgroup$ – Cherry_Developer Mar 4 '15 at 1:21
  • 1
    $\begingroup$ Yeah, that's basically the story. It sounds like you came pretty close to those conclusions, though maybe missed how to articulate it. $\endgroup$ – BaronVT Mar 4 '15 at 2:20

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