Evaluate \displaystyle\lim{x \to 0}\dfrac{\sqrt{x+9}-3}{x}.

\tfrac{1}{6} Multiply by the conjugate: \dfrac{(\sqrt{x+9}-3)(\sqrt{x+9}+3)}{x(\sqrt{x+9}+3)} = \dfrac{x}{x(\sqrt{x+9}+3)} = \dfrac{1}{\sqrt{x+9}+3}. At x=0: 1/(3+3) = 1/6.

Determining Limits Using Algebraic Manipulation

AP Calculus AB· difficulty 3/5

\displaystyle\lim_{x \to 0}\dfrac{\sqrt{x+9}-3}{x}

A
$\tfrac{1}{6}$
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B
$\tfrac{1}{3}$
C
$0$
D
$Does not exist$

Explanation

\dfrac{(\sqrt{x+9}-3)(\sqrt{x+9}+3)}{x(\sqrt{x+9}+3)} = \dfrac{x}{x(\sqrt{x+9}+3)} = \dfrac{1}{\sqrt{x+9}+3}

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