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$\lim\limits_{x \to -9} \frac{x+9}{\sqrt{x+9}}$

I know the answer is $0$ from looking at the graph but I want to know how to solve this algebraically. I tried to use L'hopital's rule but that doesn't get rid of the radical in the denominator. Then I tried to multiply the top and bottom by $\sqrt{x+9}$ and after trying to cancel stuff out I still ended up with an $x+9$ in the denominator.
The I tried to split up the limit by taking the limit of the numerator and putting that over the limit of the denominator and then put the limit in the denominator inside the radical but I would still get $0$ in the denominator.

What method am I suppose to use to solve this?

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Hint: $\frac{u}{\sqrt{u}} = \sqrt{u}$.

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$$\frac{x+9}{\sqrt{x+9}}=\frac{(x+9)^1}{(x+9)^{\frac{1}{2}}}=(x+9)^{1-\frac{1}{2}}=(x+9)^{\frac{1}{2}}$$

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Enhance

$$\frac{{x + 9}}{{\sqrt {x + 9} }} = \frac{{\left( {x + 9} \right)\sqrt {x + 9} }}{{\sqrt {x + 9} \sqrt {x + 9} }} = \frac{{\left( {x + 9} \right)\sqrt {x + 9} }}{{\left( {x + 9} \right)}} = \sqrt {x + 9} $$

Then calculate limit.

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