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Chapter 1: Problem 27

Find the limits. $$ \lim _{x \rightarrow 2^{+}} \frac{1}{|2-x|} $$

Short Answer

Expert verified
The limit is \( +\infty \).

Step by step solution

01

Analyzing the Limit Definition

We're tasked with finding the limit of a function as x approaches 2 from the right, denoted as \( x \to 2^+ \). The function given is \( \frac{1}{|2-x|} \). First, understand that as \( x \to 2^+ \), \( 2-x \) will be a small positive number since \( x > 2 \).
02

Simplifying the Absolute Value Expression

Since \( x \to 2^+ \) implies \( x > 2 \), the absolute value expression \(|2-x|\) simplifies to \(x-2\). This means our limit expression becomes \( \lim_{x \to 2^+} \frac{1}{x-2} \).
03

Evaluating the Limit

Now we evaluate the limit \( \lim_{x \to 2^+} \frac{1}{x-2} \). As \( x \to 2^+ \), the expression \( x-2 \) approaches 0 from the positive side. Thus, \( \frac{1}{x-2} \) approaches infinity because we are dividing 1 by a number that is getting very close to 0 from the positive side.

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Key Concepts

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

One-Sided Limits
In calculus, limits help us understand how a function behaves near a point. Sometimes, we specifically look at how the function behaves as we approach a point from one side only. This is called a **one-sided limit**.
When we say \( \lim_{x \to 2^+} \), we mean we are looking at the limit as \( x \) approaches 2 from values that are greater than 2. This is not about getting to 2 from both directions, just from the right.
Understanding one-sided limits helps us examine behaviors near boundaries or edges of intervals. For instance:
  • If approaching from the right, we note it as \( x \to a^+ \).
  • If approaching from the left, it鈥檚 noted as \( x \to a^- \).
This kind of limit is crucial for understanding complex behavior in graphs where the function might not be defined exactly at that point.
Functions
Functions are basic in mathematics. They take an input and provide an output based on a rule. In the limit we are considering, our function is \( \frac{1}{|2-x|} \).
The key action for this function is division by the absolute value of another expression. Absolute values ensure numbers are non-negative, which affects behavior as we approach specific points.
Here鈥檚 how functions generally operate:
  • They can involve simple operations like addition and subtraction.
  • Functions can also include more complex operations like division and absolute values - as seen in this problem.
Understanding how to simplify functions, like transforming \(|2-x|\) into \(x-2\) when \(x > 2\), makes evaluating limits easier.
Infinity in Limits
Infinity often appears in limits when values grow very large or small. In the context of our problem, as \( x \to 2^+ \), the denominator \( x-2 \) shrinks toward zero.
As the denominator gets extremely small but stays positive, the value of the function \( \frac{1}{x-2} \) becomes very large - heading towards infinity.
Here鈥檚 what happens with infinity in limits:
  • When the denominator approaches zero, the fraction grows large.
  • If coming from the positive side, this means it goes toward positive infinity.
  • This behavior illustrates the kind of things limits help us to approximate and understand without direct computation.
Recognizing when limits approach infinity is significant because it tells us about the unbounded nature of functions at particular points.

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Most popular questions from this chapter

Find values of \(x,\) if any, at which \(f\) is not continuous. $$ f(x)=\frac{x+2}{x^{2}-4} $$

Find the limits. $$ \lim _{h \rightarrow 0} \frac{\sin h}{2 h} $$

(a) Use the Intermediate-Value Theorem to show that the equation \(x+\sin x=1\) has at least one solution in the interval \([0, \pi / 6] .\) (b) Show graphically that there is exactly one solution in the interval. (c) Approximate the solution to three decimal places.

What is wrong with the following "proof" that \(\lim _{x \rightarrow 0}[(\sin 2 x) / x]=1 ?\) since \(\lim _{x \rightarrow 0}(\sin 2 x-x)=\lim _{x \rightarrow 0} \sin 2 x-\lim _{x \rightarrow 0} x=0-0=0\) if \(x\) is close to \(0,\) then \(\sin 2 x-x \approx 0\) or, equivalently, \(\sin 2 x \approx x .\) Dividing both sides of this approximate equality by \(x\) yields \((\sin 2 x) / x \approx 1 .\) That is, \(\lim _{x \rightarrow 0}[(\sin 2 x) / x]=1\)

Find values of \(x,\) if any, at which \(f\) is not continuous. $$ f(x)=\frac{2 x+1}{4 x^{2}+4 x+5} $$
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