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Chapter 2: Problem 31

Let \(f(1)=1\) and \(f(-1)=1\) and \(f(x)=\left(x^{2}-x\right) /\left(x^{2}-1\right)\) otherwise. Decide whether \(f\) is continuous at (a) \(x=1\) (b) \(x=0\) (c) \(x=-1\).

Short Answer

Expert verified
f is not continuous at x=1 and x=-1, but continuous at x=0.

Step by step solution

01

Determine Continuity at x=1

To check if the function \( f \) is continuous at \( x=1 \), we need to verify if \( \lim_{{x \to 1}} f(x) = f(1) \). Given \( f(1) = 1 \).Calculate \( \lim_{{x \to 1}} \frac{{x^2 - x}}{{x^2 - 1}} \):1. Factorize the numerator and denominator: * Numerator: \( x(x - 1) \) * Denominator: \( (x - 1)(x + 1) \)2. Cancel the \( x - 1 \) term: \[ \lim_{{x \to 1}} \frac{x}{{x + 1}} \]3. Evaluate the limit: \[ \lim_{{x \to 1}} \frac{1}{2} = \frac{1}{2} \]Since \( \lim_{{x \to 1}} f(x) = \frac{1}{2} eq f(1) = 1 \), \( f \) is not continuous at \( x=1 \).
02

Determine Continuity at x=0

To check if \( f \) is continuous at \( x=0 \), we need to compute \( \lim_{{x \to 0}} f(x) \) and see if it matches \( f(0) \). Note that \( f(0) \) isn't defined separately, so use the formula.Calculate \( \lim_{{x \to 0}} \frac{{x^2 - x}}{{x^2 - 1}} \):1. Substitute directly (since 0 doesn't cancel): * Numerator: \( 0^2 - 0 = 0 \) * Denominator: \( 0^2 - 1 = -1 \)2. The limit evaluates to 0.Therefore, \( f(0) = \frac{0}{-1} = 0 \) and \( \lim_{{x \to 0}} f(x) = 0 \), so \( f \) is continuous at \( x=0 \).
03

Determine Continuity at x=-1

To determine if \( f \) is continuous at \( x = -1 \), check if \( \lim_{{x \to -1}} f(x) = f(-1) \). Given \( f(-1) = 1 \).Calculate \( \lim_{{x \to -1}} \frac{{x^2 - x}}{{x^2 - 1}} \):1. Factor as in previous steps: * Numerator: \( x(x - 1) \) * Denominator: \( (x - 1)(x + 1) \)2. Cancel the \( x - 1 \) term (after removing indeterminate form): \[ \lim_{{x \to -1}} \frac{x}{{x + 1}} \]3. Evaluate the limit: \[ \lim_{{x \to -1}} \frac{-1}{0} \]The value \( \lim_{{x \to -1}} \frac{x}{x+1} \) leads to a division by zero, indicating a discontinuity.Since the limit doesn't exist, \( f \) is not continuous at \( x=-1 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit
The concept of the limit of a function is crucial in determining continuity. The limit at a point indicates where the function is heading as it approaches that point. For example, consider a function approaching a value of 1 as it gets closer to a certain point on the x-axis. This behavior is analyzed using
  • Direct Substitution: Plugging the point directly into the function.
  • Factorization: Simplifying the expression to avoid indeterminate forms.
  • Cancellation: Removing common factors to find a simpler form for evaluation.
Limit calculation tells us if the function at a certain point matches the function's value at that point within its domain鈥攅ssential for understanding continuous behavior.
Discontinuity
Discontinuity in a function means there is a break, jump, or hole at a certain point. It occurs when the limit of a function as it approaches a point is not equal to the function's value there, or when the limit itself does not exist. For example, if you have a limit evaluation that results in dividing by zero, like in the case of the limit as x approaches -1
  • If the limit produces an undefined result, like division by zero, the function is discontinuous at that point.
  • When a function doesn't approach the same value from both sides, it indicates a jump discontinuity.
Understanding discontinuity helps in identifying the behavior of a function and any points where it fails to be seamless.
Factorization
Factorization is a helpful technique in simplifying complex expressions, making limits easier to evaluate. By rewriting polynomials as products of their simpler components, we can often cancel out problematic factors:
  • Numerator and Denominator: Breaking them into factors exposes common elements.
  • Cancellation: Removing common factors cancels out indeterminate forms.
  • Simplifying Limits: Allows for direct computation once the indeterminate form is resolved.
In our function example, factorizing both parts and cancelling helped simplify the expression, making it more straightforward to analyze limits and continuity.
Piecewise Function
A piecewise function is defined by different rules or expressions over different intervals. For example, our function given is defined as a separate value at specific points, such as f(1)=1 and f(-1)=1, and another formula elsewhere. Understanding piecewise functions assists in evaluating continuity and limits effectively:
  • Defined at Specific Points: They can have different values explicitly stated for certain x values.
  • Continuous Assessment: Ensure the function's formula matches its value at each segment's boundaries.
  • Handling Discontinuities: Piecewise definitions might naturally include discontinuities due to different pieces.
Through such a definition, we gain more precision in assessing continuity at critical points like x=1 and x=-1.

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Most popular questions from this chapter

At a point where \(d y / d x=0\), what is special about the graph of \(y(x)\) ? Test case: \(y=x^{2}\).

Find the derivatives of the functions in exercise. Even if \(n\) is negative or a fraction, the derivative of \(x^{n}\) is \(n x^{n-1}\). $$ 1+x+\frac{1}{2} x^{2}+\frac{1}{6} x^{3}+\frac{1}{24} x^{4} $$

Complete \((x+h)^{5}=x^{5}+\) ____. What are the binomial coefficients \(\left(\begin{array}{l}5 \\ 1\end{array}\right)\) and \(\left(\begin{array}{l}5 \\\ 2\end{array}\right)\) and \(\left(\begin{array}{l}5 \\ 3\end{array}\right) ?\)

Find the limits if they exist. An \(\varepsilon-\delta\) test is not required. $$ \lim _{t \rightarrow 2} \frac{t+3}{t^{2}-2} $$

Construct your own \(f(x)\) with these discontinuities at \(x=1\). Limit for \(x \rightarrow 1^{+},\) no limit for \(x \rightarrow 1^{-}\)
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