Math Problem Statement

Let f be a function defined on (-π/2, π/2) as f(x) = [cos(π/2 + |x|)]/[sin(π/2 - |x|)]. Then, we have four options to consider regarding the continuity, differentiability, and limits at x = 0.

Solution

The problem is asking about the function f(x)f(x), defined as:

f(x)=cos(π2+x)sin(π2x),f(x) = \frac{\cos\left(\frac{\pi}{2} + |x|\right)}{\sin\left(\frac{\pi}{2} - |x|\right)},

where f(x)f(x) is given on the interval (π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right). We need to analyze the continuity, differentiability, and limit properties at x=0x = 0, and verify which of the options (A), (B), (C), or (D) are correct.

Step-by-step analysis:

  1. Continuity at x=0x = 0:

    • For continuity, we check whether the left-hand limit, right-hand limit, and the function value at x=0x = 0 are the same.
    • Since f(x)f(x) involves absolute values, it affects the behavior of the function differently on either side of x=0x = 0, so we need to calculate the limit as x0+x \to 0^+ and x0x \to 0^-.
  2. Differentiability at x=0x = 0:

    • Even if the function is continuous at x=0x = 0, we need to verify if it is differentiable by calculating the derivative at x=0x = 0.
    • The definition of the derivative can be used to check if f(0)f'(0) exists.
  3. Checking the limit limx0+f(x)f(0)x\lim_{x \to 0^+} \frac{f(x) - f(0)}{x}:

    • This limit will help us verify the validity of option (C), whether it equals 1-1.

Now, let me go ahead and calculate these steps.### Results of the calculations:

  1. Continuity at x=0x = 0:

    • The limits from the right (limx0+f(x)\lim_{x \to 0^+} f(x)) and from the left (limx0f(x)\lim_{x \to 0^-} f(x)) are both 0. Since the limits are equal, the function is continuous at x=0x = 0.
  2. Differentiability at x=0x = 0:

    • The derivative f(0)f'(0) does not exist (nan\text{nan}), meaning the function is not differentiable at x=0x = 0.
  3. Limit in option (C):

    • The limit limx0+f(x)f(0)x\lim_{x \to 0^+} \frac{f(x) - f(0)}{x} evaluates to 1-1, confirming that option (C) is correct.

Conclusion:

  • (A) is false, because f(x)f(x) is continuous at x=0x = 0.
  • (B) is true, as the function is continuous but not differentiable at x=0x = 0.
  • (C) is true, as the limit limx0+f(x)f(0)x=1\lim_{x \to 0^+} \frac{f(x) - f(0)}{x} = -1.
  • (D) is false, since f(0)f'(0) does not exist.

Would you like further details or explanations?

Here are five related questions you might explore:

  1. How is differentiability related to continuity at a point?
  2. What conditions must be satisfied for a function to be differentiable at a point?
  3. How can we graph the function f(x)f(x) near x=0x = 0 to visualize continuity and differentiability?
  4. What role does the absolute value play in determining the limits and behavior of the function?
  5. How would the results change if the function was defined differently around x=0x = 0?

Tip: When dealing with limits and absolute values, consider splitting the analysis into cases for positive and negative values of xx.

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Math Problem Analysis

Mathematical Concepts

Continuity
Differentiability
Limits

Formulas

f(x) = [cos(π/2 + |x|)]/[sin(π/2 - |x|)]
Limit definition: lim x->0 (f(x) - f(0)) / x

Theorems

Continuity Theorem
Definition of Derivative
Limit Theorem

Suitable Grade Level

College (Calculus Level)