Problem

Find the Asymptotes x/( square root of 9x^2+1)

The question asks to determine the asymptotes of the function f(x) = x / √(9x^2 + 1). Asymptotes are lines to which the graph of a function approaches but never touches. The task involves analyzing the behavior of the function as x approaches infinity (positive or negative) to identify any horizontal or vertical asymptotes, and possibly any oblique (slant) asymptotes if they exist. This may include simplifying the function, considering limits, and using knowledge of the properties of square roots and rational functions.

$\frac{x}{\sqrt{9 x^{2} + 1}}$

Answer

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Solution:

Step:1 Determine the values that cause $\frac{x}{\sqrt{9 x^{2} + 1}}$ to be undefined. The domain is all real numbers since the denominator never equals zero.

Step:2 Identify the vertical asymptotes by finding where the function is not continuous. There are no vertical asymptotes in this case.

Step:3 To find the horizontal asymptote, calculate $\lim_{x \to \infty} \frac{x}{\sqrt{9 x^{2} + 1}}$.

Step:3.1 Normalize the expression by dividing the top and bottom by $x$, the highest degree of $x$ in the denominator: $\lim_{x \to \infty} \frac{\frac{x}{x}}{\sqrt{\frac{9 x^{2}}{x^{2}} + \frac{1}{x^{2}}}}$.

Step:3.2 Proceed to find the limit.

Step:3.2.1 Simplify by removing the $x$ term: $\lim_{x \to \infty} \frac{1}{\sqrt{\frac{9 x^{2}}{x^{2}} + \frac{1}{x^{2}}}}$.

Step:3.2.2 Eliminate the $x^{2}$ terms.

Step:3.2.2.1 Simplify the expression: $\lim_{x \to \infty} \frac{1}{\sqrt{9 + \frac{1}{x^{2}}}}$.

Step:3.2.2.2 Simplify further by dividing $9$ by $1$.

Step:3.2.3 Apply the Quotient Rule for limits as $x$ approaches infinity: $\frac{\lim_{x \to \infty} 1}{\lim_{x \to \infty} \sqrt{9 + \frac{1}{x^{2}}}}$.

Step:3.2.4 The limit of a constant is the constant itself.

Step:3.2.5 Place the limit inside the square root.

Step:3.2.6 Use the Sum of Limits Rule: $\frac{1}{\sqrt{\lim_{x \to \infty} 9 + \lim_{x \to \infty} \frac{1}{x^{2}}}}$.

Step:3.2.7 The limit of a constant is the constant itself.

Step:3.3 As the denominator grows without bound, $\frac{1}{x^{2}}$ approaches zero: $\frac{1}{\sqrt{9 + 0}}$.

Step:3.4 Finalize the simplification.

Step:3.4.1 Combine $9$ and $0$.

Step:3.4.2 Express $9$ as $3^{2}$.

Step:3.4.3 Extract terms from under the radical: $\frac{1}{3}$.

Step:4 Now, calculate $\lim_{x \to -\infty} \frac{x}{\sqrt{9 x^{2} + 1}}$ for the horizontal asymptote.

Step:4.1 Normalize the expression by dividing by $x$: $\lim_{x \to -\infty} \frac{\frac{x}{x}}{-\sqrt{\frac{9 x^{2}}{x^{2}} + \frac{1}{x^{2}}}}$.

Step:4.2 Proceed to find the limit.

Step:4.2.1 Simplify by removing the $x$ term: $\lim_{x \to -\infty} \frac{1}{- \sqrt{\frac{9 x^{2}}{x^{2}} + \frac{1}{x^{2}}}}$.

Step:4.2.2 Eliminate the $x^{2}$ terms.

Step:4.2.2.1 Simplify the expression: $\lim_{x \to -\infty} \frac{1}{- \sqrt{9 + \frac{1}{x^{2}}}}$.

Step:4.2.2.2 Simplify further by dividing $9$ by $1$.

Step:4.2.3 Apply the Quotient Rule for limits as $x$ approaches negative infinity: $\frac{\lim_{x \to -\infty} 1}{\lim_{x \to -\infty} - \sqrt{9 + \frac{1}{x^{2}}}}$.

Step:4.2.4 The limit of a constant is the constant itself.

Step:4.2.5 Factor out the constant term $-1$ from the limit.

Step:4.2.6 Place the limit inside the square root.

Step:4.2.7 Use the Sum of Limits Rule: $\frac{1}{- \sqrt{\lim_{x \to -\infty} 9 + \lim_{x \to -\infty} \frac{1}{x^{2}}}}$.

Step:4.2.8 The limit of a constant is the constant itself.

Step:4.3 As the denominator grows without bound, $\frac{1}{x^{2}}$ approaches zero: $\frac{1}{- \sqrt{9 + 0}}$.

Step:4.4 Finalize the simplification.

Step:4.4.1 Combine $1$ and $-1$.

Step:4.4.1.1 Express $1$ as $-1(-1)$.

Step:4.4.1.2 Move the negative sign in front of the fraction.

Step:4.4.2 Simplify the denominator.

Step:4.4.2.1 Combine $9$ and $0$.

Step:4.4.2.2 Express $9$ as $3^{2}$.

Step:4.4.2.3 Extract terms from under the radical: $- \frac{1}{3}$.

Step:5 The horizontal asymptotes are $y = \frac{1}{3}$ and $y = - \frac{1}{3}$.

Step:6 Attempt to find oblique asymptotes using polynomial division, which is not applicable here due to the radical in the expression. No oblique asymptotes can be found.

Step:7 The complete set of asymptotes is as follows:

No Vertical Asymptotes Horizontal Asymptotes: $y = \frac{1}{3}$ and $y = - \frac{1}{3}$ No Oblique Asymptotes

Knowledge Notes:

To find the asymptotes of a function, the following knowledge points are relevant:

  1. Domain of a Function: The set of all possible input values (x-values) for which the function is defined. A function is undefined when the denominator is zero or when taking the square root of a negative number in the real number system.

  2. Vertical Asymptotes: These occur at values of x where the function approaches infinity or negative infinity. They are found by setting the denominator equal to zero and solving for x, except when the numerator also becomes zero at the same point, which may indicate a hole instead.

  3. Horizontal Asymptotes: These are horizontal lines that the graph of a function approaches as x goes to positive or negative infinity. They are found by calculating the limits of the function as x approaches infinity or negative infinity.

  4. Limits: The value that a function approaches as the input approaches some value. Limits are fundamental to calculus and are used to define continuity, derivatives, and integrals.

  5. Limit Laws: These include the Sum of Limits Rule, the Quotient Rule, and others that allow us to break down complex limits into simpler components.

  6. Oblique Asymptotes: These occur when the degree of the numerator is exactly one greater than the degree of the denominator. They are found using polynomial long division or synthetic division, but are not present when the function includes a radical that cannot be removed.

  7. Simplifying Expressions: Involves canceling common factors, rationalizing denominators, and combining like terms to make expressions easier to work with, especially when evaluating limits.

Understanding these concepts is crucial for analyzing the behavior of functions and finding their asymptotes.

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