AP Calculus ABmediummcq1 pt

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

\frac{1}{3}

\frac{1}{6}

0

D.Does not exist

Explanation

Core Concept

Multiply by the conjugate: $\frac{\sqrt{x+9}-3}{x}\cdot\frac{\sqrt{x+9}+3}{\sqrt{x+9}+3}=\frac{x}{x(\sqrt{x+9}+3)}=\frac{1}{\sqrt{x+9}+3}$ . As $x\to 0$ this approaches $\frac{1}{3+3}=\frac{1}{6}$ .

Correct Answer

B $\frac{1}{6}$

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lim⁡x→0x+9−3x=\lim_{x\to 0}\dfrac{\sqrt{x+9}-3}{x}=limx→0​xx+9​−3​=

Explanation

Core Concept

Correct Answer

More Unit 1: Limits and Continuity practice questions

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