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Question

Mathematics Question on Limits

Let f (x) = 118−x2 \frac {1} {\sqrt {18-x^2}}, ten value o f lim⁡x→3(f(x)−f(3)x−3)\lim_{x \to 3} (\frac {f(x)-f(3)} {x-3})

A

0

B

−19-\frac {1} {9}

C

−13-\frac {1} {3}

D

19\frac {1} {9}

Answer

19\frac {1} {9}

Explanation

Solution

lim⁡x→3(18−x2)−12−(13)x−3=lim⁡x→33−18−x23(x−3)18−x2\lim _{x\to 3} \frac{\left(18 -x^{2}\right)^{-\frac{1}{2} }- \left(\frac{1}{3}\right)}{x-3} =\lim _{x\to 3} \frac{3 -\sqrt{18 -x^{2}}}{3\left(x-3\right) \sqrt{18 -x^{2}} } = lim⁡x→3[9−(18−x2)3(x−3)18−x2(3+18−x2)]\lim _{x\to 3} \left[\frac{9-\left(18 -x^{2}\right)}{3\left(x-3\right) \sqrt{18 -x^{2}}\left(3+\sqrt{18 -x^{2}}\right)}\right] = lim⁡x→3x2−93(x−3)18−x2[3+18−x2]\lim _{x\to 3} \frac{x^{ 2} -9}{3\left(x-3\right) \sqrt{18 -x^{2} } \left[3+\sqrt{18 -x^{2}}\right]} = lim⁡x→3x+3318−x2[3+18−x2]\lim _{x\to 3} \frac{x+3}{3\sqrt{18-x^{2}} \left[3+\sqrt{18 -x^{2}}\right] } = 63(3)(3+3)=19\frac{6}{3\left(3\right)\left(3+3\right)} =\frac{1}{9}