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Chapter 2: Problem 24

Continuity at a point Determine whether the following functions are continuous at a. Use the continuity checklist to justify your answer. $$f(x)=\left\\{\begin{array}{ll}\frac{x^{2}+x}{x+1} & \text { if } x \neq-1 \\\2 & \text { if } x=-1\end{array} ; a=-1\right.$$

Short Answer

Expert verified
Provide an explanation. No, the function \(f(x)\) is not continuous at \(x=-1\). Although the function is defined at \(x=-1\) and the limit exists as \(x\) approaches \(-1\), the limit of the function does not equal the value of the function at \(x=-1\). This means it fails to satisfy the third condition for continuity, and is therefore not continuous at \(x=-1\).

Step by step solution

01

Check if the function is defined at \(x=a\)#

We are given that \(f(x) = 2\) if \(x=-1\). Since there is a value for the function at \(x=-1\), the function is defined at this point. So, condition 1 is satisfied.#
02

Calculate the limit of the function as x approaches a#

In order to determine the limit of the function as \(x\) approaches \(-1\), we'll consider the limit to be \(f(-1)\), when the value of \(x\) is not equal to \(-1\). $$f(x)=\frac{x^{2}+x}{x+1}, \text{ for } x\neq -1$$ To find the limit of this function as \(x\) approaches \(-1\), we can simplify the expression by factoring the numerator and cancelling the common terms: $$\lim _{x \rightarrow-1} f(x)=\lim _{x \rightarrow-1} \frac{x^{2}+x}{x+1} =\lim _{x \rightarrow-1} \frac{x(x+1)}{x+1}$$ Now we can cancel the common term \((x+1)\) from the numerator and denominator, leaving us with: $$\lim _{x \rightarrow-1} x$$ Now that the expression is simplified, we can substitute \(x=-1\) to find the limit: $$\lim _{x \rightarrow-1} x = -1$$ So the limit of the function as \(x\) approaches \(-1\) exists, and \(\lim _{x\rightarrow -1} f(x) = -1\). This satisfies condition 2.#
03

Compare the limit of the function as x approaches a with the value of the function at x=a#

Lastly, we need to determine if the calculated limit is equal to the value of the function at \(x=-1\): $$\lim _{x \rightarrow-1} f(x) = -1 \neq f(-1) = 2$$ Since the limit of the function as \(x\) approaches \(-1\) does not equal the value of the function at \(x=-1\), condition 3 is not satisfied.#
04

Conclusion#

Using the continuity checklist, we conclude that the function \(f(x)\) is not continuous at \(x=-1\), as it does not satisfy all three conditions for continuity.

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Key Concepts

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

Limits
Limits are a fundamental concept in calculus, representing the value that a function approaches as the input approaches a certain point. When we find the limit of a function as x approaches a specific value, we're essentially trying to predict the function's behavior near that value.

Calculating limits involves:
  • Understanding the behavior of the function near the point of interest.
  • Substituting values that get increasingly close to the point into the function.
  • Using algebraic techniques like factoring to simplify expressions.
In our original exercise, the process involves factoring the expression \((x^2 + x)\) to \((x(x+1))\), canceling out the common term from the denominator, and then substituting the value to find the limit. By doing this, we're able to determine that \(lim_{x \rightarrow -1} f(x) = -1\).

Understanding limits is crucial because they help us determine the continuity of functions and are a stepping stone to learning derivatives and integrals.
Discontinuous Functions
A discontinuous function is a function that is not continuous at one or more points in its domain. For a function to be continuous at a point, three conditions must be met:
  • The function must be defined at that point.
  • The limit of the function as it approaches the point must exist.
  • The limit of the function as it approaches the point must equal the function's value at that point.
In our exercise, when examining the function at \(x = -1\), although the function is defined and the limit exists, the limit \(\lim_{x \rightarrow -1} f(x) = -1\) does not equal \(f(-1) = 2\).

Due to this mismatch, the function is discontinuous at this point. Recognizing points of discontinuity is vital in calculus because they often indicate a change or a point where certain calculus techniques, like differentiation, may not apply.
Factoring
Factoring is a mathematical technique used to simplify complex expressions, often making it easier to work with limits and identify points of discontinuity. It involves breaking down a complex expression into simpler, multiplied parts.

In the given exercise, we encounter the expression \(\frac{x^2 + x}{x + 1}\). The numerator can be factored as \(x(x + 1)\). By factoring, we can cancel the common term \(x + 1\) from both the numerator and the denominator. This simplification allows us to safely evaluate the limit without directly substituting into a problematic expression.

Factoring serves multiple purposes in calculus:
  • Simplifies expressions making them easier to analyze.
  • Helps identify removable discontinuities, where terms cancel out to remove undefined points.
  • Provides insight into the behavior of polynomial expressions.
A strong grasp of factoring techniques will greatly aid in tackling various calculus problems, leading to more robust problem-solving skills.

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Most popular questions from this chapter

Sketching functions a. Sketch the graph of a function that is not continuous at \(1,\) but is defined at 1 b. Sketch the graph of a function that is not continuous at \(1,\) but has a limit at 1

Violation of the Intermediate Value Theorem? Let \(f(x)=\frac{|x|}{x} .\) Then \(f(-2)=-1\) and \(f(2)=1 .\) Therefore \(f(-2)<0

Continuity of compositions a. Find functions \(f\) and \(g\) such that each function is continuous at 0 but \(f \circ g\) is not continuous at 0 b. Explain why examples satisfying part (a) do not contradict Theorem 2.11

Find the following limits or state that they do not exist. Assume \(a, b, c,\) and k are fixed real numbers. $$\lim _{x \rightarrow 3} \frac{x-3}{|x-3|}$$

One-sided limits Let $$g(x)=\left\\{\begin{array}{ll}5 x-15 & \text { if } x<4 \\\\\sqrt{6 x+1} & \text { if } x \geq 4\end{array}\right.$$ Compute the following limits or state that they do not exist. a. \(\lim _{x \rightarrow 4} g(x)\) b. \(\lim _{x \rightarrow 4^{+}} g(x)\) c. \(\lim _{x \rightarrow 4} g(x)\)
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