Let f(x)={[(15 x)/(x^(3)+3x)," for "x!=0],[k," for "x=0]:} f is continuous for all real numbers. What is the value of k ? Choose 1 answer: (A) 5 (B) 0 (C) 8 (D) 3

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Let  f(x)={[(15 x)/(x^(3)+3x)," for "x!=0],[k," for "x=0]:}  f is continuous for all real numbers. What is the value of  k ? Choose 1 answer: (A) 5 (B) 0 (C) 8 (D) 3

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Define Limit of f(x): To determine the value of k that makes f(x) continuous at x = 0, we need to find the limit of f(x) as x approaches 0 from both the left and the right. If the limit exists and is equal to k, then f(x) is continuous at x = 0.Factor Out x: We calculate the limit of the function as x approaches 0: $$\lim_{x \to 0} \frac{15x}{x^3 + 3x}$$ We can factor out an x from the denominator: $$\lim_{x \to 0} \frac{15x}{x(x^2 + 3)}$$ Now we can simplify the fraction by canceling out the common x term: $$\lim_{x \to 0} \frac{15}{x^2 + 3}$$Evaluate Limit at x=0: As x approaches 0, the x^2 term also approaches 0, so we can evaluate the limit by substituting x with 0: $$\lim_{x \to 0} \frac{15}{0 + 3} = \frac{15}{3} = 5$$Determine Value of k: Since the limit of f(x) as x approaches 0 is 5, for f to be continuous at x = 0, k must be equal to 5. This is because the definition of continuity at a point requires that the limit of the function as it approaches the point from both sides is equal to the function's value at that point.

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