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Chapter 1: Problem 1

Given that $$ \lim _{x \rightarrow a} f(x)=2, \quad \lim _{x \rightarrow a} g(x)=-4, \quad \lim _{x \rightarrow a} h(x)=0 $$ find the limits. (a) \(\lim _{x \rightarrow a}[f(x)+2 g(x)]\) (b) \(\lim _{x \rightarrow a}[h(x)-3 g(x)+1]\) (c) \(\lim _{x \rightarrow a}[f(x) g(x)]\) (d) \(\lim _{x \rightarrow a}[g(x)]^{2}\) (e) \(\lim _{x \rightarrow a} \sqrt[3]{6+f(x)}\) (f) \(\lim _{x \rightarrow a} \frac{2}{g(x)}\)

Short Answer

Expert verified
(a) -6, (b) 13, (c) -8, (d) 16, (e) 2, (f) -1/2.

Step by step solution

01

Understanding Limit Rules

When we are given the limits of functions as \( x \) approaches a specific value, we can find the limits of combinations of these functions by using limit properties. There are several key properties we will use: linearity, product, power, and constant factor rules.
02

Solve Part (a)

To find \( \lim_{x \to a} [f(x) + 2g(x)] \), use the linearity property of limits. This property states that the limit of a sum or difference is the sum or difference of their limits. Using given limits, we have:\[ \lim_{x \to a} [f(x) + 2g(x)] = \lim_{x \to a} f(x) + 2 \cdot \lim_{x \to a} g(x) \]\[ = 2 + 2 \cdot (-4) = 2 - 8 = -6 \]
03

Solve Part (b)

For \( \lim_{x \to a} [h(x) - 3g(x) + 1] \), apply the linearity property:\[ \lim_{x \to a} [h(x) - 3g(x) + 1] = \lim_{x \to a} h(x) - 3 \cdot \lim_{x \to a} g(x) + \lim_{x \to a} 1 \]Since the limit of a constant is the constant itself, we have:\[ = 0 - 3 \cdot (-4) + 1 = 0 + 12 + 1 = 13 \]
04

Solve Part (c)

To find \( \lim_{x \to a} [f(x) g(x)] \), use the product property of limits:\[ \lim_{x \to a} [f(x) g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \]\[ = 2 \cdot (-4) = -8 \]
05

Solve Part (d)

To compute \( \lim_{x \to a} [g(x)]^2 \), use the power property of limits:\[ \lim_{x \to a} [g(x)]^2 = (\lim_{x \to a} g(x))^2 \]\[ = (-4)^2 = 16 \]
06

Solve Part (e)

Here, you are to find \( \lim_{x \to a} \sqrt[3]{6+f(x)} \). Using the continuity of the cube root and linearity:\[ \lim_{x \to a} \sqrt[3]{6+f(x)} = \sqrt[3]{6 + \lim_{x \to a} f(x)} \]\[ = \sqrt[3]{6 + 2} = \sqrt[3]{8} = 2 \]
07

Solve Part (f)

For \( \lim_{x \to a} \frac{2}{g(x)} \), we must use the rule for the limit of a quotient, assuming the limit of the denominator is non-zero:\[ \lim_{x \to a} \frac{2}{g(x)} = \frac{2}{\lim_{x \to a} g(x)} \]\[ = \frac{2}{-4} = -\frac{1}{2} \]

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Key Concepts

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

Understanding Limit Properties
The concept of limits in calculus helps us understand how functions behave as they approach a certain point. Limit properties are rules that allow us to manipulate and simplify expressions involving limits. These rules include:
  • Linearity Property: You can take limits over sums or differences by considering the limit of each part separately and then summing or subtracting the results.
  • Product Property: The limit of a product is the product of the limits, provided the limits exist independently.
  • Power Property: The limit of an expression raised to a power can be found by taking the limit of the base and then raising it to the given power.
  • Constant Factor Rule: You can factor out a constant when finding the limit of an expression.
By understanding and applying these properties, it becomes easier to solve complex limit problems without evaluating them point by point.
Continuity in Limits
Continuity ensures that small changes in the input of a function result in small changes in the output. A function is continuous at a point if its limit at that point equals the function's value there.
  • For continuous functions, you can evaluate limits directly by substituting the point into the function.
  • Discontinuities occur when there are sudden jumps, infinite behaviors, or undefined points in a function.
  • Continuity plays a crucial role in determining the behavior of complex expressions composed of multiple continuous functions.
Thus, recognizing continuity enables smoother problem-solving, especially when using limit properties like linearity and product properties.
Product Property of Limits
When determining the limit of a product of functions, we can use the product property. The product property states that:
  • The limit of the product of two functions is equal to the product of their individual limits: \[ \ \lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \\]
  • This property is particularly useful when each function鈥檚 limit is known and finite.
  • It simplifies finding limits of multiplication expressions by allowing the evaluation of each term separately.
This approach provides an efficient way to tackle products in limits by reducing complexity into smaller, manageable pieces.
Linearity Property of Limits
Linearity is one of the fundamental properties utilized when working with limits. This property indicates:
  • The limit of a sum is the sum of the limits:\[ \ \lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) \\]
  • The limit of a difference is the difference of the limits:\[ \ \lim_{x \to a} [f(x) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x) \\]
  • Constant factors can also be multiplied with limits straightforwardly.
Using the linearity property simplifies solving complex limit problems involving sums and differences, as it breaks them down into easier calculations by evaluating each part separately.

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Most popular questions from this chapter

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

The population \(p\) of the United States (in millions) in year \(t\) can be modeled by the function $$p(t)=\frac{525}{1+1.1 e^{-0.02225(t-1990)}}$$ (a) Based on this model, what was the U.S. population in \(1990 ?\) (b) Plot \(p\) versus \(t\) for the 200 -year period from 1950 to 2150 (c) By evaluating an appropriate limit, show that the graph of \(p\) versus \(t\) has a horizontal asymptote \(p=c\) for an appropriate constant \(c\) (d) What is the significance of the constant \(c\) in part (c) for the population predicted by this model?

(a) Explain why the following calculation is incorrect. $$ \begin{aligned} \lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{x^{2}}\right) &=\lim _{x \rightarrow 0^{+}} \frac{1}{x}-\lim _{x \rightarrow 0^{+}} \frac{1}{x^{2}} \\ &=+\infty-(+\infty)=0 \end{aligned} $$ (b) Show that \(\lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)=-\infty\)

(a) Explain informally why $$\lim _{x \rightarrow 0^{-}}\left(\frac{1}{x}+\frac{1}{x^{2}}\right)=+\infty$$ (b) Verify the limit in part (a) algebraically.

The notion of an asymptote can be extended to include curves as well as lines. Specifically, we say that curves \(y=f(x)\) and \(y=g(x)\) are asymptotic as \(x \rightarrow+\infty\) provided $$ \lim _{x \rightarrow+\infty}[f(x)-g(x)]=0 $$ and are asymptotic as \(x \rightarrow-\infty\) provided $$ \lim _{x \rightarrow-\infty}[f(x)-g(x)]=0 $$ In these exercises, determine a simpler function \(g(x)\) such that \(y=f(x)\) is asymptotic to \(y=g(x)\) as \(x \rightarrow+\infty\) or \(x \rightarrow-\infty\) Use a graphing utility to generate the graphs of \(y=f(x)\) and \(y=g(x)\) and identify all vertical asymptotes. $$f(x)=\frac{x^{5}-x^{3}+3}{x^{2}-1}$$
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