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91Ó°ÊÓ

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

Short Answer

Expert verified
Answer: The limit does not exist.

Step by step solution

01

Identify when x < 3 and x > 3

We will evaluate the expression separately for the cases when x approaches 3 from the left side (x < 3) and when x approaches 3 from the right side (x > 3).
02

Evaluate the expression for x < 3

When x approaches 3 from the left side (x < 3), then (x - 3) is negative and the absolute value function |x - 3| equals -(x - 3), so the expression becomes: $$\lim _{x \rightarrow 3^{-}} \frac{x-3}{|x-3|} = \lim _{x \rightarrow 3^{-}} \frac{x-3}{-(x-3)} = \lim _{x \rightarrow 3^{-}} -1$$
03

Evaluate the expression for x > 3

When x approaches 3 from the right side (x > 3), then (x - 3) is positive and the absolute value function |x - 3| equals (x - 3), so the expression becomes: $$\lim _{x \rightarrow 3^{+}} \frac{x-3}{|x-3|} = \lim _{x \rightarrow 3^{+}} \frac{x-3}{x-3} = \lim _{x \rightarrow 3^{+}} 1$$
04

Find the limit if the values match, otherwise state that they do not exist

Since the limits from the left and right sides are different, the overall limit does not exist, so we can state: $$\lim _{x \rightarrow 3} \frac{x-3}{|x-3|} \text{ does not exist.}$$

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Key Concepts

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

One-Sided Limits
When we talk about one-sided limits, we are essentially investigating how a function behaves as our variable approaches a particular point from one direction only. Now, let's take a closer look at the exercise given. We are examining the expression \( \lim _{x \rightarrow 3} \frac{x-3}{|x-3|} \) by splitting it into two cases: one from the left (as \( x \) approaches 3 from values less than 3) and one from the right (as \( x \) approaches 3 from values greater than 3).

First, the left-side limit (denoted as \( x \rightarrow 3^{-} \)), means we approach our point of interest slightly below 3. In this setup, \( x - 3 \) is a negative value. Therefore, the expression becomes \( \lim _{x \rightarrow 3^{-}} \frac{x-3}{-(x-3)} = -1 \). Next, on the right-side limit (denoted as \( x \rightarrow 3^{+} \)), we approach from values greater than 3. Here, \( x - 3 \) is positive, turning our expression into \( \lim _{x \rightarrow 3^{+}} \frac{x-3}{x-3} = 1 \). The essence of one-sided limits is capturing this distinct behavior from both directions.

Ultimately, identifying how a function approaches a point from either direction helps us determine overall continuity and pinpoint any drastic changes at that point.
Absolute Value
The concept of absolute value is crucial while analyzing limits and discontinuities. Absolute value, written as \(|x|\), is the distance of a number from zero on the number line, always non-negative. For a function \( |x-3| \), it represents the distance of \( x \) from 3 and dictates how we handle different cases near \( x = 3 \).

In the exercise \( \lim _{x \rightarrow 3} \frac{x-3}{|x-3|} \), understanding absolute value becomes key when handling expressions approaching from both sides. Because \( x - 3 \) is negative when \( x < 3 \), the absolute value \( |x-3| = -(x-3) \). Conversely, when \( x > 3 \), \( x - 3 \) is positive, so \( |x-3| = x-3 \). This changes how the fraction \( \frac{x-3}{|x-3|} \) behaves, turning into \(-1\) when approaching from the left, and \(1\) from the right.

The use of absolute values shapes our expressions and helps us understand the boundary behavior in functions, especially when differences appear depending on the side from which the limit is approached.
Limit Does Not Exist
Discovering when a limit does not exist involves noticing the inconsistency between the left-side and right-side limits. Limits are a way to see how a function behaves as it approaches a certain value. When these behavioral outcomes differ on each side of our target point, the overall limit does not exist.

In our exercise, the calculation \( \lim _{x \rightarrow 3} \frac{x-3}{|x-3|} \) led to two distinct results: \(-1\) from the left and \(1\) from the right. Since these results diverge, they cannot form a single unified result for the limit at \( x = 3 \).

This discrepancy means the limit is undefined at this point. Hence, we conclude that \( \lim _{x \rightarrow 3} \frac{x-3}{|x-3|} \) does not exist. Recognizing when limits do not exist is essential, particularly for identifying discontinuities and alerting us to potential breakpoints in a function's graph. Understanding such discontinuities can be crucial for analyzing comprehensive mathematical relationships and ensuring function continuity.

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Most popular questions from this chapter

Assume you invest \(\$ 250\) at the end of each year for 10 years at an annual interest rate of \(r .\) The amount of money in your account after 10 years is given by \(A(r)=\frac{250\left((1+r)^{10}-1\right)}{r} .\) Assume your goal is to have \(\$ 3500\) in your account after 10 years. a. Show that there is an interest rate \(r\) in the interval \((0.01,0.10)-\) between \(1 \%\) and \(10 \%-\) that allows you to reach your financial goal. b. Use a calculator to estimate the interest rate required to reach your financial goal.

Continuity of \(\sin x\) and \(\cos x\). a. Use the identity \(\sin (a+h)=\sin a \cos h+\cos a \sin h\) with the fact that \(\lim _{x \rightarrow 0} \sin x=0\) to prove that \(\lim _{x \rightarrow a} \sin x=\sin a\) thereby establishing that \(\sin x\) is continuous for all \(x .\) (Hint: Let \(h=x-a\) so that \(x=a+h\) and note that \(h \rightarrow 0\) as \(x \rightarrow a .)\) b. Use the identity \(\cos (a+h)=\cos a \cos h-\sin a \sin h\) with the fact that \(\lim _{x \rightarrow 0} \cos x=1\) to prove that \(\lim _{x \rightarrow q} \cos x=\cos a\).

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