Let f(x)=ln2(x−1)−x.Select the correct description of the one-sided limits of f at x=2.Choose 1 answer:(A)limx→2+f(x)=+∞ and limx→2−f(x)=+∞(B)limx→2+f(x)=+∞ and limx→2−f(x)=−∞(C)limx→2+f(x)=−∞ and limx→2−f(x)=+∞(D)limx→2+f(x)=−∞ and limx→2−f(x)=−∞
Q. Let f(x)=ln2(x−1)−x.Select the correct description of the one-sided limits of f at x=2.Choose 1 answer:(A)limx→2+f(x)=+∞ and limx→2−f(x)=+∞(B)limx→2+f(x)=+∞ and limx→2−f(x)=−∞(C)limx→2+f(x)=−∞ and limx→2−f(x)=+∞(D)limx→2+f(x)=−∞ and limx→2−f(x)=−∞
Analyze Right Approach: Analyze the behavior of the function as x approaches 2 from the right (x→2+).As x approaches 2 from the right, the numerator −x approaches −2. The denominator ln2(x−1) approaches ln2(1), which is 0. Since the denominator approaches 0 and the numerator approaches a negative value, the function 21 approaches negative infinity.
Analyze Left Approach: Analyze the behavior of the function as x approaches 2 from the left (x→2−).As x approaches 2 from the left, the numerator −x still approaches −2. However, we must consider the behavior of the natural logarithm function ln(x−1) as x approaches 2 from the left. The natural logarithm of a number close to 20 from the left is negative, and as x approaches 2, ln(x−1) approaches 24 which is 25. Since the denominator is the square of ln(x−1), it will be a small positive number. Therefore, the function 27 approaches negative infinity as well.
Choose Correct Answer: Choose the correct answer based on the analysis.From Step 1 and Step 2, we have determined that both one-sided limits as x approaches 2 lead to negative infinity. Therefore, the correct answer is:(D) {limx→2+f(x)=−∞ and limx→2−f(x)=−∞}
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