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Chapter 1: Problem 21

Find the limit. $$ \lim _{x \rightarrow-4}(x+3)^{2} $$

Short Answer

Expert verified
The limit of the function as \(x\) approaches -4 is 1.

Step by step solution

01

Identify the value \(x\) is approaching

According to the question, we need to evaluate \(\lim _{x \rightarrow -4}(x+3)^{2}\), which means \(x\) is approaching -4.
02

Substitute the value of \(x\) into the function (Direct substitution as it is a continuous function)

We substitute \(x = -4\) into the function expression giving us \((-4+3)^{2}\). This simplifies to \((-1)^{2}\).
03

Evaluate the expression

Now simplify \((-1)^{2}\) to find the limit which gives us a result of 1.

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Key Concepts

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

Limits
The concept of a limit is fundamental in calculus. It allows us to understand what value a function is approaching as the input approaches a certain point. In our exercise, we are interested in finding \( \lim \_{x \rightarrow -4}(x+3)^{2} \), which means we want to know what \( (x+3)^{2} \) becomes as \( x \) gets very close to -4.
To think about limits:
  • Consider the behavior of the function's output as the input approaches a particular value.
  • If the outputs are getting closer and closer to a specific number, that number is the limit.
Understanding limits is crucial because they help us assess functions that may not be easy to evaluate directly due to complexities or indeterminate forms, such as \( \frac{0}{0} \). In our case, however, evaluating the limit is straightforward because the expression \( (x+3)^{2} \) is simple.
Continuous Functions
A function is termed continuous if it is smooth at every point within its domain, meaning there are no gaps, jumps, or breaks.
The function \( f(x) = (x+3)^{2} \) in the exercise is a continuous function. This property is important because:
  • It means that you can apply the operation of taking limits by direct substitution, making calculations much simpler.
  • As \( x \) approaches -4, you only need to focus on plugging in the value to determine the output.
For functions like polynomials, rational functions (with non-zero denominators at the point of interest), sine, cosine, etc., continuity is often guaranteed. Recognizing continuous functions saves a lot of time and makes calculus concepts more approachable.
Direct Substitution
Direct substitution is a method used to find the limit of a function at a given point, especially when the function is continuous. It involves:
  • Directly substituting the value of \( x \) into the function.
  • No additional manipulation is required if the function is continuous at that point.
In our exercise, since \( f(x) = (x+3)^{2} \) is continuous, evaluating \( \lim \_{x \rightarrow -4}(x+3)^{2} \) involves simply substituting \( -4 \) into \( (x+3)^{2} \).
This gives us \( (-4+3)^{2} \). Each step breaks down further by simplifying to \( (-1)^{2} \), which results in \( 1 \).
Direct substitution is a powerful tool in calculus since it simplifies the process of finding limits for continuous functions.
Function Evaluation
Function evaluation refers to the process of determining the output of a function for a particular input. It is often one of the final steps when solving limit problems involving continuous functions.To evaluate a function:
  • Take the function expression and replace the variable \( x \) with the given value.
  • Perform any arithmetic operations necessary to simplify the expression.
In the exercise, evaluation is performed by substituting \( x = -4 \) into \( (x+3)^{2} \) leading to \( (-4+3)^{2} \).
Simplifying further gives \( (-1)^{2} = 1 \), showing the limit as \( x \to -4 \) is indeed \( 1 \). This process solidifies how direct substitution within continuous functions leads directly to function evaluation, making complex problems simpler.

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Most popular questions from this chapter

Use a graphing utility to graph the function. Use the graph to determine any \(x\) -values at which the function is not continuous. $$ h(x)=\frac{1}{x^{2}-x-2} $$

(a) Let \(f_{1}(x)\) and \(f_{2}(x)\) be continuous on the closed interval \([a, b]\). If \(f_{1}(a)f_{2}(b)\), prove that there exists \(c\) between \(a\) and \(b\) such that \(f_{1}(c)=f_{2}(c)\). (b) Show that there exists \(c\) in \(\left[0, \frac{\pi}{2}\right]\) such that \(\cos x=x\). Use a graphing utility to approximate \(c\) to three decimal places.

Verify that the Intermediate Value Theorem applies to the indicated interval and find the value of \(c\) guaranteed by the theorem. $$ f(x)=x^{3}-x^{2}+x-2, \quad[0,3], \quad f(c)=4 $$

Discuss the continuity of the composite function \(h(x)=f(g(x))\). $$ \begin{aligned} &f(x)=\frac{1}{\sqrt{x}} \\ &g(x)=x-1 \end{aligned} $$

Use a graphing utility to graph the function. Use the graph to determine any \(x\) -values at which the function is not continuous. $$ g(x)=\left\\{\begin{array}{ll} 2 x-4, & x \leq 3 \\ x^{2}-2 x, & x>3 \end{array}\right. $$
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