Evaluate ModifyingBelow lim With x right arrow infinity f left parenthesis x right parenthesis and ModifyingBelow lim With x right arrow minus infinity f left parenthesis x right parenthesis for the following function. Then give the horizontal​ asymptote(s) of f​ (if any).
f left parenthesis x right parenthesis equals StartFraction x cubed plus 4 Over 2 x cubed plus StartRoot 16 x Superscript 6 Baseline plus 1 EndRoot EndFraction
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Part 1
Evaluate ModifyingBelow lim With x right arrow infinity f left parenthesis x right parenthesis. Select the correct choice​ and, if​ necessary, fill in the answer box to complete your choice.
A.
ModifyingBelow lim With x right arrow infinity f left parenthesis x right parenthesisequals
  
one sixth
 ​(Type an integer or a simplified​ fraction.)
Your answer is correct.B.
The limit does not exist and is neither infinity nor minusinfinity.
Part 2
Evaluate ModifyingBelow lim With x right arrow minus infinity f left parenthesis x right parenthesis. Select the correct choice​ and, if​ necessary, fill in the answer box to complete your choice.
A.
ModifyingBelow lim With x right arrow minus infinity f left parenthesis x right parenthesisequals
  
enter your response here ​(Type an integer or a simplified​ fraction.)
B.
The limit does not exist and is neither infinity nor minusinfinity.

To evaluate the limits of the function \( f(x) = \frac{x^3 + 4}{2x^3 + \sqrt{16x^6 + 1}} \) as \( x \) approaches \( \infty \) and \( -\infty \), let's first simplify the expression for large values of \( x \).

### Simplification of the function:
1. **Factor out \( x^3 \) from the numerator and denominator:**
   \[
   f(x) = \frac{x^3(1 + \frac{4}{x^3})}{x^3 \left( 2 + \frac{\sqrt{16x^6 + 1}}{x^3} \right)}
   \]
2. **Simplify the expression:**
   \[
   f(x) = \frac{1 + \frac{4}{x^3}}{2 + \frac{\sqrt{16x^6 + 1}}{x^3}}
   \]

### Analyzing the limit as \( x \to \infty \):

For large \( x \), the term \( \frac{4}{x^3} \) becomes negligible, and the term \( \sqrt{16x^6 + 1} \) can be approximated as:
\[
\sqrt{16x^6 + 1} \approx \sqrt{16x^6} = 4x^3
\]
So, the expression for \( f(x) \) simplifies to:
\[
f(x) \approx \frac{1}{2 + \frac{4x^3}{x^3}} = \frac{1}{2 + 4} = \frac{1}{6}
\]

Thus:
\[
\lim_{x \to \infty} f(x) = \frac{1}{6}
\]

### Analyzing the limit as \( x \to -\infty \):

When \( x \) is negative and large in magnitude, the approximation \( \sqrt{16x^6 + 1} \approx 4x^3 \) still holds (since \( x^6 \) is positive), and the function simplifies in the same way as for positive large \( x \):
\[
f(x) \approx \frac{1}{6}
\]

Thus:
\[
\lim_{x \to -\infty} f(x) = \frac{1}{6}
\]

### Horizontal Asymptotes:
Since both limits \( \lim_{x \to \infty} f(x) \) and \( \lim_{x \to -\infty} f(x) \) equal \( \frac{1}{6} \), the horizontal asymptote of the function \( f(x) \) is \( y = \frac{1}{6} \).

**Summary:**
- \( \lim_{x \to \infty} f(x) = \frac{1}{6} \)
- \( \lim_{x \to -\infty} f(x) = \frac{1}{6} \)
- The horizontal asymptote is \( y = \frac{1}{6} \).

Would you like further details on this process or have any additional questions? Here are some related questions to consider:

1. What are the vertical asymptotes of a rational function?
2. How do you evaluate limits at infinity for more complex functions?
3. What happens to the horizontal asymptotes when the degrees of the numerator and denominator differ?
4. Can a function have more than one horizontal asymptote?
5. What is the significance of horizontal asymptotes in graphing functions?

**Tip:** When dealing with limits involving square roots, always consider factoring or approximating the square root expression for large values of \( x \).

Learn how to evaluate limits as x approaches infinity and minus infinity for a rational function. Understand the concept of horizontal asymptotes and their significance in graphing functions.

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