Problem 56
Question
Use a table of values to evaluate the following limits as \(x\) decreases without bound. $$\lim _{x \rightarrow-\infty} \frac{7 x^{3}}{5 x^{2}+3 x}$$
Step-by-Step Solution
Verified Answer
The limit is \(-\infty\).
1Step 1: Simplify the Expression
We start by analyzing the given function \( f(x) = \frac{7x^3}{5x^2 + 3x} \). We notice that as \( x \) becomes very large or very small, the terms with the highest powers will dominate the behavior of the function.
2Step 2: Identify Dominant Terms
In the numerator \( 7x^3 \), the \( x^3 \) term is the most significant. In the denominator, \( 5x^2 \) is the leading term. This indicates that as \( x \to -\infty \), these terms will heavily influence the value of the fraction.
3Step 3: Approximate the Expression for Large |x|
As \( x \to -\infty \), the function can be approximated by considering only the dominant terms: \( \frac{7x^3}{5x^2} \). Simplifying this gives \( \frac{7x}{5} \).
4Step 4: Analyze Behavior as x Approaches Negative Infinity
Since we have \( \frac{7x}{5} \), as \( x \to -\infty \), the expression \( \frac{7x}{5} \) will decrease without bound because of the negative sign of \( x \).
5Step 5: Conclude the Limit
Therefore, the limit of the function as \( x \to -\infty \) is \(-\infty\).
Key Concepts
Table of ValuesDominant TermsApproximating FunctionsNegative Infinity
Table of Values
Sometimes, it's helpful to use a table of values to evaluate limits, especially when dealing with functions that may seem complex at first glance. In our example, the function is given by \( \frac{7x^3}{5x^2 + 3x} \). By plugging different values of \( x \) into this function, we can see how it behaves as \( x \) approaches negative infinity.
To create a table of values, choose several large negative numbers for \( x \), such as -10, -100, and -1000. Calculate the corresponding \( f(x) \) values for each:
\[\begin{align*}x = -10, & \quad f(x) \approx \frac{7(-10)^3}{5(-10)^2 + 3(-10)} \x = -100, & \quad f(x) \approx \frac{7(-100)^3}{5(-100)^2 + 3(-100)} \x = -1000, & \quad f(x) \approx \frac{7(-1000)^3}{5(-1000)^2 + 3(-1000)} \\end{align*}\]
Notice how these calculations yield increasingly negative results, reinforcing that as \( x \to -\infty \), the function heads towards \(-\infty\). Observe these trends and use them to conclude the behavior of the function.
To create a table of values, choose several large negative numbers for \( x \), such as -10, -100, and -1000. Calculate the corresponding \( f(x) \) values for each:
\[\begin{align*}x = -10, & \quad f(x) \approx \frac{7(-10)^3}{5(-10)^2 + 3(-10)} \x = -100, & \quad f(x) \approx \frac{7(-100)^3}{5(-100)^2 + 3(-100)} \x = -1000, & \quad f(x) \approx \frac{7(-1000)^3}{5(-1000)^2 + 3(-1000)} \\end{align*}\]
Notice how these calculations yield increasingly negative results, reinforcing that as \( x \to -\infty \), the function heads towards \(-\infty\). Observe these trends and use them to conclude the behavior of the function.
Dominant Terms
When evaluating limits, especially as \( x \to \pm \infty \), identifying dominant terms is crucial. These are the terms in polynomials that have the highest power and significantly influence the function's behavior.
In our function \( \frac{7x^3}{5x^2 + 3x} \), the term \( 7x^3 \) in the numerator is dominant because it has the highest power of \( x \) compared to both the numerator and denominator. In the denominator, \( 5x^2 \) is the leading term because it has the highest degree among the terms present.
By focusing only on these dominant terms, we can simplify the function considerably, especially when \( x \) takes on very large or very small values. This simplification helps us approximate the function more easily and understand its limiting behavior.
In our function \( \frac{7x^3}{5x^2 + 3x} \), the term \( 7x^3 \) in the numerator is dominant because it has the highest power of \( x \) compared to both the numerator and denominator. In the denominator, \( 5x^2 \) is the leading term because it has the highest degree among the terms present.
By focusing only on these dominant terms, we can simplify the function considerably, especially when \( x \) takes on very large or very small values. This simplification helps us approximate the function more easily and understand its limiting behavior.
Approximating Functions
Approximating functions using dominant terms is a key strategy in evaluating limits as \( x \to \infty \) or \( -\infty \). This approach simplifies complex functions, allowing us to focus on the most significant terms and ignore less impactful ones for large values of \( |x| \).
For the given function \( \frac{7x^3}{5x^2 + 3x} \), we approximate it by using the dominant terms, resulting in \( \frac{7x^3}{5x^2} \). This simplifies further to \( \frac{7x}{5} \).
Through this simplification, we see that \( f(x) \) behaves linearly as \( x \to -\infty \). The approximated function \( \frac{7x}{5} \) gives us a clear view of how \( f(x) \) behaves without getting lost in less significant terms. By understanding this approach, you’ll be better equipped to tackle similar problems.
For the given function \( \frac{7x^3}{5x^2 + 3x} \), we approximate it by using the dominant terms, resulting in \( \frac{7x^3}{5x^2} \). This simplifies further to \( \frac{7x}{5} \).
Through this simplification, we see that \( f(x) \) behaves linearly as \( x \to -\infty \). The approximated function \( \frac{7x}{5} \) gives us a clear view of how \( f(x) \) behaves without getting lost in less significant terms. By understanding this approach, you’ll be better equipped to tackle similar problems.
Negative Infinity
Exploring behavior as \( x \to -\infty \) involves understanding what happens when \( x \) decreases without bound. In this context, mathematical terms like 'dominant' become pivotal, where only the most impactful terms matter.
For the function \( \frac{7x}{5} \), as \( x \) approaches \(-\infty\), \( x \) itself pulls the expression to negative infinity because multiplying a negative number by a positive fraction like \( \frac{7}{5} \) continues to decrease the result.
The negative sign of \( x \) ensures that the function value continues to drop, emphasizing its downward trend. Understanding this concept is essential because it helps you grasp why the limit of the original function as \( x \to -\infty \) is indeed \(-\infty\). It's not merely about the algebra but understanding how the behavior of \( x \) influences the outcome.
For the function \( \frac{7x}{5} \), as \( x \) approaches \(-\infty\), \( x \) itself pulls the expression to negative infinity because multiplying a negative number by a positive fraction like \( \frac{7}{5} \) continues to decrease the result.
The negative sign of \( x \) ensures that the function value continues to drop, emphasizing its downward trend. Understanding this concept is essential because it helps you grasp why the limit of the original function as \( x \to -\infty \) is indeed \(-\infty\). It's not merely about the algebra but understanding how the behavior of \( x \) influences the outcome.
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