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Chapter 3: Problem 60

Find the limit, if it exists. $$ \lim _{x \rightarrow-\infty} \frac{7 x^{5}+x-9}{6 x+x^{3}} $$

Short Answer

Expert verified
The limit does not exist, as it approaches infinity.

Step by step solution

01

Identify the Leading Terms

To find the limit as \( x \rightarrow -\infty \), identify the terms with the highest powers of \(x\) in the numerator and the denominator. In this case, the leading term in the numerator is \(7x^5\) and the leading term in the denominator is \(x^3\).
02

Simplify the Expression

Divide every term in the numerator and the denominator by \(x^3\), which is the highest power of \(x\) in the denominator. This will give us: \[ \frac{ \frac{7x^5}{x^3} + \frac{x}{x^3} - \frac{9}{x^3} }{ \frac{6x}{x^3} + \frac{x^3}{x^3} } = \frac{7x^2 + \frac{1}{x^2} - \frac{9}{x^3}}{\frac{6}{x^2} + 1} \]
03

Evaluate the Limit for Each Term

As \( x \rightarrow -\infty \), \( 7x^2 \rightarrow \infty \), \( \frac{1}{x^2} \rightarrow 0 \), \( \frac{9}{x^3} \rightarrow 0 \), and \( \frac{6}{x^2} \rightarrow 0 \). Substituting these into our simplified expression, we get: \[ \lim_{x \rightarrow -\infty} \frac{7x^2 + 0 - 0}{0 + 1} = \lim_{x \rightarrow -\infty} 7x^2 = \infty \]

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Key Concepts

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

Leading Terms in Limits
When evaluating limits, especially for expressions involving large values of x, it's crucial to identify the leading terms. Leading terms are the terms with the highest power of the variable, as they often dominate the behavior of the function at extreme values. In our exercise, we have the expression: \[ \lim _{x \rightarrow -\infty} \frac{7 x^{5}+x-9}{6 x+x^{3}} \]Here, the leading term in the numerator is \(7x^{5}\) and in the denominator, it's \(x^{3}\). Since limits at infinity often focus on the most significant terms, we simplify the problem by concentrating on these leading terms.
Dividing by Highest Power
Once we've identified the leading terms, the next step is to divide every term in the equation by the highest power found in the denominator. For our exercise, we've noted that the highest power in the denominator is \(x^{3}\). By dividing each term by \(x^{3}\), we simplify the original expression: \[ \frac{7x^5 + x - 9}{6x + x^3} \rightarrow \frac{ \frac{7x^5}{x^3} + \frac{x}{x^3} - \frac{9}{x^3} }{ \frac{6x}{x^3} + \frac{x^3}{x^3} } = \frac{7x^2 + \frac{1}{x^2} - \frac{9}{x^3}}{\frac{6}{x^2} + 1} \]This simplification makes it easier to evaluate the behavior of the numerator and denominator as \( x \) approaches negative infinity.
Evaluating Limits at Infinity
With our simplified expression, the next step is to evaluate the limit for each individual term as \( x \rightarrow -\infty \). Let's break it down:
  • The term \(7x^{2}\) tends to \( \infty \).
  • The terms \(\frac{1}{x^{2}}\) and \(\frac{9}{x^{3}}\) both tend to 0, because as \( x \) becomes very large in magnitude, the fractions become very small.
  • The term \(\frac{6}{x^{2}}\) also tends to 0 for the same reason.
  • The constant term 1 remains as 1.
So, we have: \[ \lim_{x \rightarrow -\infty} \frac{7x^2 + 0 - 0}{0 + 1} = \lim_{x \rightarrow -\infty} 7x^2 = \infty \]Thus, the limit of the given expression as \( x \rightarrow -\infty \) is \( \infty \).

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