Let h(x)=(sqrt(x+12)-3)/(x+3) when x!=-3. h is continuous for all x -12. Find h(-3). Choose 1 answer: (A) -(1)/(9) (B) (1)/(9) (c) -(1)/(6) (D) (1)/(6)

Question

Let  h(x)=(sqrt(x+12)-3)/(x+3) when  x!=-3.  h is continuous for all  x > -12. Find  h(-3). Choose 1 answer: (A)  -(1)/(9) (B)  (1)/(9) (c)  -(1)/(6) (D)  (1)/(6)

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Introduction: We are asked to find the value of h(-3), but we cannot directly substitute x = -3 into the function because the denominator becomes zero, which would result in an undefined expression. To find the limit as x approaches -3, we need to simplify the expression to eliminate the zero in the denominator.Rationalizing the Numerator: We notice that the numerator contains a square root, which suggests that we might be able to simplify the expression by rationalizing the numerator. To do this, we multiply the numerator and the denominator by the conjugate of the numerator, which is (sqrt(x+12)+3).Expression Simplification: The expression becomes: h(x) = [(sqrt(x+12)-3)(sqrt(x+12)+3)] / [(x+3)(sqrt(x+12)+3)] This simplifies to: h(x) = [(x+12) - 9] / [(x+3)(sqrt(x+12)+3)]Cancellation of Terms: Further simplification gives us: h(x) = (x + 12 - 9) / [(x+3)(sqrt(x+12)+3)] h(x) = (x + 3) / [(x+3)(sqrt(x+12)+3)]Substituting x = -3: We can now cancel out the (x+3) term in the numerator and the denominator, as long as x is not equal to -3. This is valid because we are looking for the limit as x approaches -3, not the value at x = -3. h(x) = 1 / (sqrt(x+12)+3)Final Simplification: Now we can safely substitute x = -3 into the simplified expression to find the limit as x approaches -3: h(-3) = 1 / (sqrt(-3+12)+3) h(-3) = 1 / (sqrt(9)+3)Simplifying further, we get: h(-3) = 1 / (3+3) h(-3) = 1 / 6

Let $h(x)=\frac{\sqrt{x+12}-3}{x+3}$ when $x \neq-3$ . $\newline$ $h$ is continuous for all x>-12 . $\newline$ Find $h(-3)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $-\frac{1}{9}$ $\newline$ (B) $\frac{1}{9}$ $\newline$ (c) $-\frac{1}{6}$ $\newline$ (D) $\frac{1}{6}$

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