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f(x)=(sqrt(3x-15)-3)/(x-8) We want to find lim_(x rarr8)f(x). What happens when we use direct substitution? Choose 1 answer: (A) The limit exists, and we found it! (B) The limit doesn't exist (probably an asymptote). (C) The result is indeterminate.

f(x)=3x−15−3x−8 f(x)=\frac{\sqrt{3 x-15}-3}{x-8} \newlineWe want to find lim⁡x→8f(x) \lim _{x \rightarrow 8} f(x) .\newlineWhat happens when we use direct substitution?\newlineChoose 11 answer:\newline(A) The limit exists, and we found it!\newline(B) The limit doesn't exist (probably an asymptote).\newline(C) The result is indeterminate.

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Q. f(x)=3x−15−3x−8 f(x)=\frac{\sqrt{3 x-15}-3}{x-8} \newlineWe want to find lim⁡x→8f(x) \lim _{x \rightarrow 8} f(x) .\newlineWhat happens when we use direct substitution?\newlineChoose 11 answer:\newline(A) The limit exists, and we found it!\newline(B) The limit doesn't exist (probably an asymptote).\newline(C) The result is indeterminate.
  1. Direct Substitution: Let's first try direct substitution of x=8x = 8 into the function f(x)f(x) to see what happens.\newlinef(x)=3x−15−3x−8f(x) = \frac{\sqrt{3x - 15} - 3}{x - 8}\newlinef(8)=3⋅8−15−38−8f(8) = \frac{\sqrt{3\cdot8 - 15} - 3}{8 - 8}\newlinef(8)=24−15−30f(8) = \frac{\sqrt{24 - 15} - 3}{0}\newlinef(8)=9−30f(8) = \frac{\sqrt{9} - 3}{0}\newlinef(8)=3−30f(8) = \frac{3 - 3}{0}\newlinef(8)=00f(8) = \frac{0}{0}
  2. Rationalize the Numerator: Since we have an indeterminate form, we can try to simplify the expression to eliminate the indeterminate form. One common technique is to rationalize the numerator.\newlineLet's multiply the numerator and the denominator by the conjugate of the numerator.\newlineThe conjugate of 3x−15\sqrt{3x - 15} - 33 is 3x−15\sqrt{3x - 15} + 33.\newlinef(x)=(3x−15−3)∗(3x−15+3)(x−8)∗(3x−15+3)f(x) = \frac{(\sqrt{3x - 15} - 3) * (\sqrt{3x - 15} + 3)}{(x - 8) * (\sqrt{3x - 15} + 3)}
  3. Simplify the Numerator: Now, let's perform the multiplication in the numerator, which is a difference of squares.\newline(3x−15−3)∗(3x−15+3)=(3x−15)−(3)2(\sqrt{3x - 15} - 3) * (\sqrt{3x - 15} + 3) = (3x - 15) - (3)^2\newline(3x−15)−9=3x−24(3x - 15) - 9 = 3x - 24
  4. Factor Out Common Factor: We can now substitute the simplified numerator back into the function.\newlinef(x)=3x−24(x−8)⋅(3x−15+3)f(x) = \frac{3x - 24}{(x - 8) \cdot (\sqrt{3x - 15} + 3)}\newlineNotice that the numerator now has a factor of 33 that can be factored out.\newlinef(x)=3(x−8)(x−8)⋅(3x−15+3)f(x) = \frac{3(x - 8)}{(x - 8) \cdot (\sqrt{3x - 15} + 3)}
  5. Final Substitution: We can see that (x−8)(x - 8) is a common factor in the numerator and the denominator, so we can cancel it out.\newlinef(x)=3(3x−15+3)f(x) = \frac{3}{(\sqrt{3x - 15} + 3)}\newlineNow, let's try direct substitution again with x=8x = 8.\newlinef(8)=3(3⋅8−15+3)f(8) = \frac{3}{(\sqrt{3\cdot8 - 15} + 3)}\newlinef(8)=3(24−15+3)f(8) = \frac{3}{(\sqrt{24 - 15} + 3)}\newlinef(8)=3(9+3)f(8) = \frac{3}{(\sqrt{9} + 3)}\newlinef(8)=3(3+3)f(8) = \frac{3}{(3 + 3)}\newlinef(8)=36f(8) = \frac{3}{6}\newlinef(8)=12f(8) = \frac{1}{2}

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