/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 90 True or False: If \(\lim _{x \ri... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

True or False: If \(\lim _{x \rightarrow 2} f(x)=7\), then \(\lim _{x \rightarrow 2^{+}} f(x)=7\)

Short Answer

Expert verified
True

Step by step solution

01

Understanding the Problem

We are asked to determine if the statement is true or false. The statement claims that if the two-sided limit of a function \(f(x)\) as \(x\) approaches 2 is 7, then the right-hand limit as \(x\) approaches 2 from the positive side is also 7.
02

Concept of Two-Sided Limits

The two-sided limit \(\lim_{x \to 2} f(x) = 7\) means that as \(x\) gets closer to 2 from both sides (left \(2^{-}\) and right \(2^{+}\)), the value of \(f(x)\) gets closer to 7.
03

Implication of Two-Sided Limits

For a two-sided limit to exist and equal a certain value, both the left-hand and right-hand limits must exist and equal that value. Therefore, if \(\lim_{x \to 2} f(x) = 7\), then both \(\lim_{x \to 2^{+}} f(x)\) and \(\lim_{x \to 2^{-}} f(x)\) must equal 7.
04

Conclusion

Since the condition \(\lim_{x \to 2} f(x) = 7\) implies that the right-hand limit also equals 7, the statement is true.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Two-Sided Limits
Two-sided limits help us understand the behavior of a function as the input, denoted by \(x\), gets closer to a specific point from both directions, left and right. This is expressed mathematically as \(\lim_{x \to a} f(x) = L\), where \(L\) is the limit value when \(x\) approaches the point \(a\).

When we talk about a two-sided limit existing at a certain value, we mean that no matter how small we choose a number around \(a\), the function \(f(x)\) gets closer to \(L\) from both sides.

For example:
  • If \(x\) approaches a value from the left (written as \(x \to a^{-}\)), and the function nears \(L\).
  • And if \(x\) approaches from the right (written as \(x \to a^{+}\)), and the function also nears \(L\).
Only when both conditions are satisfied do we say the two-sided limit exists and equals \(L\). Hence, in our exercise, \(\lim_{x \to 2} f(x) = 7\) implies that from both left and right sides, the values of \(f(x)\) converge to 7.
Right-Hand Limits
The right-hand limit is specifically concerned with the behavior of a function \(f(x)\) as \(x\) approaches a particular point \(a\) from the positive, or right, side. It is symbolized as \(\lim_{x \to a^{+}} f(x)\).

This means that we are only interested in how \(f(x)\) behaves when \(x\) is slightly larger than \(a\). It does not consider values of \(x\) less than \(a\).

In the context of the exercise, we explore the right-hand limit \(\lim_{x \to 2^{+}} f(x)\). In this scenario, as \(x\) approaches 2 from values greater than 2, if \(f(x)\) gets closer to 7, the right-hand limit at \(a = 2\) is 7.

This aligns with the conclusion in the step-by-step solution: if the two-sided limit \(\lim_{x \to 2} f(x)\) is 7, the right-hand limit can also be deduced to be 7, confirming the statement as true.
Limit Existence
The existence of a limit is a crucial concept in calculus, determining whether a function approaches a particular value at a given point. When seeking to establish limit existence at a point \(a\), both the right-hand and left-hand limits must meet certain criteria:
  • The right-hand limit, \(\lim_{x \to a^{+}} f(x)\), must exist.
  • The left-hand limit, \(\lim_{x \to a^{-}} f(x)\), must also exist.


  • If both of these limits converge to the same value \(L\), we can say the two-sided limit exists, equaling \(L\), thus making the overall limit exist at that point.

    In the exercise, the statement \(\lim_{x \to 2} f(x) = 7\) signifies complete limit existence because both one-sided limits (left and right) equate to the same value. This dual agreement confirms that the limit at \(x = 2\) indeed exists and affirms the answer to the question, making the given claim true.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The Chain Rule (page 155 ) states that the derivative of \(f(g(x))\) is \(f^{\prime}(g(x)) \cdot g^{\prime}(x)\). Use Carathéodory's definition of the derivative to prove the Chain Rule by giving reasons for the following steps. a. Since \(g\) is differentiable at \(x\), there is a function \(G\) that is continuous at 0 and such that \(g(x+h)-g(x)=G(h) \cdot h\), and \(G(0)=g^{\prime}(x)\). b. Since \(f\) is differentiable at \(g(x)\), there is a function \(F\) that is continuous at 0 and such that \(f(g(x)+h)-f(g(x))=F(h) \cdot h\), and \(F(0)=f^{\prime}(g(x))\) c. For the function \(f(g(x))\) we have \(f(g(x+h))-f(g(x))\) \(\quad=f(g(x)+g(x+h)-g(x))-f(g(x))\) \(\quad=f(g(x)+(g(x+h)-g(x)))-f(g(x))\) \(=F(g(x+h)-g(x)) \cdot(g(x+h)-g(x))\) \(=F(g(x+h)-g(x)) \cdot G(h) \cdot h\) Therefore, the derivative of \(f(g(x))\) is \(\begin{aligned} F(g(x+0)-g(x)) \cdot G(0) &=F(0) \cdot G(0) \\ &=f^{\prime}(g(x)) \cdot g^{\prime}(x) \end{aligned}\) as was to be proved.

If \(\lim _{x \rightarrow 2} f(x)=7\) and \(f(x)\) is continuous at \(x=2\), then \(f(2)=7\).

For a function \(f(x)\), if \(f\) is in widgets and \(x\) is in blivets, what are the units of the derivative \(f^{\prime}(x)\), widgets per blivet or blivets per widget?

Derive the Quotient Rule from the Product Rule as follows. a. Define the quotient to be a single function, $$ Q(x)=\frac{f(x)}{g(x)} $$ b. Multiply both sides by \(g(x)\) to obtain the equation \(Q(x) \cdot g(x)=f(x) .\) c. Differentiate each side, using the Product Rule on the left side. d. Solve the resulting formula for the derivative \(Q^{\prime}(x) .\) e. Replace \(Q(x)\) by \(\frac{f(x)}{g(x)}\) and show that the resulting formula for \(Q^{\prime}(x)\) is the same as the Quotient Rule. Note that in this derivation when we differentiated \(Q(x)\) we assumed that the derivative of the quotient exists, whereas in the derivation on pages \(135-136\) we proved that the derivative exists.

A study estimated how a person's social status (rated on a scale where 100 indicates the status of a college graduate) depended on years of education. Based on this study, with \(e\) years of education, a person's status is \(S(e)=0.22(e+4)^{2.1}\). Find \(S^{\prime}(12)\) and interpret your answer.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.