# Evaluate the limit and use L'Hopital's Rule if necessary.

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

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.

• No, it's just because $\sqrt x+ x<6$ when $x<9$ (note the function is increasing) – Adam Hughes Mar 4 '15 at 0:57
• $\sqrt 8 + 8 < 6$? It is the denominator that is responsible for the change in sign. – 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.
• 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. – Cherry_Developer Mar 4 '15 at 1:21