Math Problem Statement

Determine the type of discontinuity at x = 2 for the functions f(x) = (x - 2) / |x - 2| and g(x) = (√x - √2) / (x - 2).

Solution

To determine the type of discontinuity at x=2x = 2 for each of the given functions, we need to analyze their behavior at that point.

Part (a): f(x)=x2x2f(x) = \frac{x - 2}{|x - 2|}

  1. Definition Check: Notice that the function involves an absolute value in the denominator, which will affect the behavior of f(x)f(x) differently based on whether xx is greater than or less than 2.

  2. Behavior as x2+x \to 2^+:

    • When x>2x > 2, x2=x2|x - 2| = x - 2.
    • Substituting into f(x)f(x): f(x)=x2x2=1f(x) = \frac{x - 2}{x - 2} = 1
  3. Behavior as x2x \to 2^-:

    • When x<2x < 2, x2=(x2)|x - 2| = -(x - 2).
    • Substituting into f(x)f(x): f(x)=x2(x2)=1f(x) = \frac{x - 2}{-(x - 2)} = -1
  4. Conclusion:

    • As x2x \to 2 from the left, f(x)1f(x) \to -1.
    • As x2x \to 2 from the right, f(x)1f(x) \to 1.
    • Since the left-hand limit and the right-hand limit are not equal, there is a jump discontinuity at x=2x = 2 for f(x)f(x).

Part (b): g(x)=x2x2g(x) = \frac{\sqrt{x} - \sqrt{2}}{x - 2}

  1. Direct Substitution: Plugging in x=2x = 2 directly gives an indeterminate form 00\frac{0}{0}, so we’ll use L'Hôpital's Rule.

  2. Applying L'Hôpital's Rule:

    • Differentiate the numerator and the denominator with respect to xx.
      • The derivative of x2\sqrt{x} - \sqrt{2} with respect to xx is 12x\frac{1}{2\sqrt{x}}.
      • The derivative of x2x - 2 with respect to xx is 1.
    • Applying L'Hôpital's Rule: limx2g(x)=limx212x1=122\lim_{x \to 2} g(x) = \lim_{x \to 2} \frac{\frac{1}{2\sqrt{x}}}{1} = \frac{1}{2\sqrt{2}}
  3. Conclusion:

    • The limit exists and is finite, so g(x)g(x) has a removable discontinuity at x=2x = 2.
    • We could redefine g(x)g(x) at x=2x = 2 as 122\frac{1}{2\sqrt{2}} to make it continuous at that point.

Would you like a deeper breakdown of these steps or have any questions?


Here are five related questions to further explore this topic:

  1. How do different types of discontinuities (jump, removable, infinite) affect the continuity of a function?
  2. Can all discontinuities be removed by redefining the function at a point?
  3. How does the absolute value function impact the limits of functions as xx approaches a given point?
  4. When is it appropriate to use L'Hôpital's Rule for evaluating limits?
  5. How can piecewise functions be used to define functions with jump discontinuities?

Tip: For functions with absolute values, consider the behavior on either side of the point in question, as it often leads to different expressions for the limit.

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Math Problem Analysis

Mathematical Concepts

Limits
Discontinuity
Absolute Value
Square Root
L'Hôpital's Rule

Formulas

Limit of a function as x approaches a point
L'Hôpital's Rule for 0/0 indeterminate forms
Properties of absolute value and square root functions

Theorems

L'Hôpital's Rule
Definition of Jump and Removable Discontinuities

Suitable Grade Level

College Calculus