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 near x=2: Analyze the function near x=2. We have the function f(x)=ln2(x−1)−x. We need to consider the behavior of the function as x approaches 2 from the left (x→2−) and from the right (x→2+). We will look at the numerator and the denominator separately.
Numerator behavior at x=2: Consider the behavior of the numerator as x approaches 2. The numerator is −x. As x approaches 2 from either side, −x approaches −2. There is no ambiguity or complexity in the behavior of the numerator.
Denominator behavior at x=2: Consider the behavior of the denominator as x approaches 2. The denominator is ln2(x−1), which is the square of the natural logarithm of (x−1). As x approaches 2, (x−1) approaches 1. The natural logarithm of 1 is x0, so ln2(x−1) approaches x0. We need to consider the sign of the natural logarithm as x approaches 2 from the left and right to determine the sign of the denominator.
Denominator sign at x→2−: Determine the sign of the denominator as x approaches 2 from the left. As x approaches 2 from the left (x→2−), (x−1) is slightly less than 1. The natural logarithm of a number slightly less than 1 is negative, so ln(x−1) is negative. Squaring a negative number gives a positive result, so x0 is positive as x approaches 2 from the left.
Denominator sign at x→2+: Determine the sign of the denominator as x approaches 2 from the right.As x approaches 2 from the right (x→2+), (x−1) is slightly more than 1. The natural logarithm of a number slightly more than 1 is positive, so ln(x−1) is positive. Squaring a positive number gives a positive result, so x0 is positive as x approaches 2 from the right.
Combine for one-sided limits: Combine the behavior of the numerator and denominator to find the one-sided limits.Since the numerator approaches −2 and the denominator approaches 0 with a positive sign from both sides, the fractionln2(x−1)−x will approach negative infinity from both sides. This is because a negative number divided by a positive number that is getting closer and closer to zero will result in a number that is getting more and more negative.
More problems from Compare linear, exponential, and quadratic growth