Find the Asymptotes f(x)=x/( square root of 9x^2+1)
This problem is asking you to determine the lines called asymptotes for the given function f(x) = x / √(9x^2 + 1). An asymptote is a line that the graph of a function approaches but never actually reaches as the independent variable (usually x) either goes to infinity or to a certain point. There are three types of asymptotes: vertical, horizontal, and oblique (slant).
The question likely is seeking to identify any and all of these asymptotes for the provided function. To find horizontal asymptotes, you typically look at the behavior of the function as x approaches infinity or negative infinity. For vertical asymptotes, you would be looking for values of x that make the function undefined (though this function does not have any vertical asymptotes due to the square root in the denominator). For oblique asymptotes, they occur when the degree of the polynomial in the numerator is one degree higher than the denominator; they can be found using long division or other analytical methods if applicable.
The problem is essentially a test of understanding limits and the behavior of rational functions at extreme values of x, as well as knowledge of how to identify different types of asymptotic behavior.
$f \left(\right. x \left.\right) = \frac{x}{\sqrt{9 x^{2} + 1}}$
Answer
Solution:
Step 1:
Determine if the function $\frac{x}{\sqrt{9x^2 + 1}}$ is undefined for any real number. The function is defined for all real numbers.
Step 2:
Identify any vertical asymptotes by looking for points where the function is not continuous. There are no vertical asymptotes for this function.
Step 3:
To find horizontal asymptotes, calculate the limit of the function as $x$ approaches infinity.
Step 3.1:
Normalize the function by dividing the numerator and denominator by $x$, the highest power of $x$ in the denominator: $\lim_{x \to \infty} \frac{\frac{x}{x}}{\sqrt{\frac{9x^2}{x^2} + \frac{1}{x^2}}}$.
Step 3.2:
Simplify the limit expression.
Step 3.2.1:
Eliminate the common $x$ term: $\lim_{x \to \infty} \frac{1}{\sqrt{\frac{9x^2}{x^2} + \frac{1}{x^2}}}$.
Step 3.2.2:
Reduce the fraction by canceling out $x^2$: $\lim_{x \to \infty} \frac{1}{\sqrt{9 + \frac{1}{x^2}}}$.
Step 3.2.3:
Apply the limit quotient rule: $\frac{\lim_{x \to \infty} 1}{\lim_{x \to \infty} \sqrt{9 + \frac{1}{x^2}}}$.
Step 3.2.4:
The limit of a constant is the constant itself: $\frac{1}{\lim_{x \to \infty} \sqrt{9 + \frac{1}{x^2}}}$.
Step 3.2.5:
Place the limit operation inside the radical: $\frac{1}{\sqrt{\lim_{x \to \infty} 9 + \frac{1}{x^2}}}$.
Step 3.2.6:
Separate the limits: $\frac{1}{\sqrt{\lim_{x \to \infty} 9 + \lim_{x \to \infty} \frac{1}{x^2}}}$.
Step 3.2.7:
Evaluate the limit of the constant and the fraction: $\frac{1}{\sqrt{9 + 0}}$.
Step 3.3:
Simplify the expression: $\frac{1}{3}$.
Step 4:
Calculate the limit as $x$ approaches negative infinity to find any additional horizontal asymptotes.
Step 4.1:
Normalize the function: $\lim_{x \to -\infty} \frac{\frac{x}{x}}{-\sqrt{\frac{9x^2}{x^2} + \frac{1}{x^2}}}$.
Step 4.2:
Proceed with simplification as in steps 3.2.1 to 3.2.7, but considering the negative infinity limit.
Step 4.3:
The result of the limit as $x$ approaches negative infinity is $-\frac{1}{3}$.
Step 5:
The horizontal asymptotes are $y = \frac{1}{3}$ and $y = -\frac{1}{3}$.
Step 6:
Check for any oblique asymptotes by attempting polynomial division. The presence of a radical in the function precludes polynomial division, so there are no oblique asymptotes.
Step 7:
Compile the list of all asymptotes for the function: no vertical asymptotes, horizontal asymptotes at $y = \frac{1}{3}$ and $y = -\frac{1}{3}$, and no oblique asymptotes.
Knowledge Notes:
To find the asymptotes of a function, we typically follow these steps:
Vertical Asymptotes: These occur where the function is undefined, often due to division by zero. To find them, we look for values of $x$ that make the denominator zero (without also making the numerator zero).
Horizontal Asymptotes: These are found by evaluating the limit of the function as $x$ approaches positive or negative infinity. If the limit exists, it is the value of the horizontal asymptote.
Oblique Asymptotes: These occur when the degree of the numerator is exactly one higher than the degree of the denominator. We find them by performing polynomial long division.
In the given function $\frac{x}{\sqrt{9x^2 + 1}}$, there are no vertical asymptotes because the denominator never equals zero for any real number $x$. Horizontal asymptotes are found by taking the limits as $x$ approaches positive and negative infinity. Since the degree of the numerator and denominator are the same, we expect horizontal asymptotes. Oblique asymptotes are not possible in this case because the function does not have a polynomial numerator and denominator.
