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In Exercises \(47-56,\) use your intuition to decide whether the limit exists. Justify your answer by using the rigorous definition of limit. $$ \begin{array}{l} \lim _{x \rightarrow 3} f(x) \text { where } \\ \qquad f(x)=\left\\{\begin{array}{cl} x^{2} & \text { if } x \leq 0 \\ 1-x^{2} & \text { if } x>0 \end{array}\right. \end{array} $$

Short Answer

Expert verified
The limit exists and is \(-8\).

Step by step solution

01

Analyze the Function

The function \( f(x) \) is defined piecewise. For \( x \leq 0 \), it is \( f(x) = x^2 \), and for \( x > 0 \), it is \( f(x) = 1 - x^2 \). We need to find the limit as \( x \rightarrow 3 \). Observe that for values of \( x > 0 \), the function is \( f(x) = 1-x^2 \). At \( x = 3 \), this piece of the function is applicable.
02

Evaluate the Limit Directly

Since \( x \rightarrow 3 \) implies \( x > 0 \), use the function expression for \( x > 0 \). Thus, substitute \( x = 3 \) in \( 1-x^2 \) to calculate the limit:\[\lim_{x \rightarrow 3} f(x) = 1 - 3^2 = 1 - 9 = -8.\]
03

Confirm Using Rigorous Definition

According to the rigorous definition of a limit, for \( \lim_{x \to 3} f(x) = L \), for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - 3| < \delta \), then \( |f(x) - L| < \epsilon \). Here, \( L = -8 \). Substitute \( x \) in \( f(x) = 1 - x^2 \) for values close to \( x = 3 \), confirming the function outputs close to \(-8\) if \( x \) is sufficiently close to 3.
04

Conclusion on Limit Existence

Since we can find such a \( \delta \) for any \( \epsilon > 0 \) that satisfies the condition that \( |f(x) + 8| < \epsilon \) when \( x \) is close to 3, the limit exists and is \(-8\).

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Key Concepts

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

Understanding Piecewise Functions
Piecewise functions are unique in that they are defined by multiple expressions over different intervals of the domain. In this exercise, the function \( f(x) \) is defined in two parts: for \( x \leq 0 \), the expression is \( x^2 \), and for \( x > 0 \), the expression is \( 1 - x^2 \). Piecewise functions often require separate analysis in different sections of their domains.

This structure allows a function to behave differently over various intervals, like being one type of polynomial in one range and another elsewhere. When dealing with limits of piecewise functions, it is crucial to choose the correct piece for evaluation based on the limit point and any conditions like \( x > 0 \) or \( x \leq 0 \).

Understanding which part of the function to use requires recognizing where your point of interest (in this case, \( x = 3 \)) falls in the domain intervals given by the function. Since 3 is greater than 0, we use the second piece, \( 1-x^2 \), to evaluate the limit.
The Rigorous Definition of Limit
The rigorous definition of a limit is fundamental in calculus and provides a precise way of describing the behavior of functions as the input approaches a particular value. For a function \( f(x) \), the limit \( \lim_{x \to c} f(x) = L \) if, for every positive number \( \epsilon \), there exists a positive number \( \delta \) such that whenever \( 0 < |x - c| < \delta \), it follows that \( |f(x) - L| < \epsilon \).

This ensures that as \( x \) gets arbitrarily close to \( c \), the values of \( f(x) \) approach \( L \) within any desired level of accuracy. The delta-epsilon definition is a mathematical way to capture the notion that we can make \( f(x) \) as close as we like to \( L \) by making \( x \) sufficiently close to \( c \).

When analyzing piecewise functions with limits, we apply this definition specifically to the relevant piece of the function at the point we're investigating. This exercise shows that by using \( f(x) = 1 - x^2 \) and seeing \( \lim_{x \to 3} f(x) = -8 \), it's possible to confirm the closeness condition by choosing appropriate \( \delta \) as required by the definition.
Evaluating Limits
Evaluating limits is about finding where the function is heading as the input approaches a certain value. In piecewise functions, this involves determining which part of the function to use. When evaluating \( \lim_{x \rightarrow 3} f(x) \), since \( x > 0 \) when approaching 3, we apply the expression \( 1 - x^2 \).

To compute this limit directly, substitute \( x = 3 \) into the appropriate function part, giving \( 1 - 3^2 = 1 - 9 = -8 \). This provides the limit value straightforwardly, but confirmation through the rigorous definition using the epsilon-delta approach ensures accuracy by showing continuity over intervals near 3.

Successfully finding limits involves attentively examining conditions given by the function's definition, and applying the correct part to smoothly evaluate as \( x \) approaches the desired point, supported by theoretical backing like the delta-epsilon criteria to affirm the accuracy of the limit value. This double-check assures that no unexpected behavior or discontinuity at the approaching juncture falsely suggests the wrong limit.

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Most popular questions from this chapter

An assertion is made about a function \(f\) that is defined on a closed, bounded interval. If the statement is true, explain why. Otherwise, sketch a function \(f\) that shows it is false. (Note: \(|f|\) is defined by \(|f|(x)=|f(x)| .)\) If \(f^{2}\) is continuous, then \(f\) is continuous.

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In each of Exercises \(93-96,\) plot the given functions \(g, f,\) and \(h\) in a common viewing rectangle that illustrates \(f\) being pinched at the point \(c\). Determine \(\lim _{x \rightarrow c} f(x)\). $$ \begin{array}{l} g(x)=2-4|x|, f(x)=\left(2+x^{2}\right) /\left(1+x^{2}\right)+\sin (x+1 / x)+ \\ \sin (x-1 / x), h(x)=2(1+|x|) \quad c=0 \end{array} $$

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