Find the Asymptotes (4x)/( square root of 16x^2+5)

Solution:

Step 1:

Determine the values for which the function $\frac{4x}{\sqrt{16x^2+5}}$ is not defined. Since the denominator is a square root, it's defined for all $x$ in the real numbers.

Step 2:

Identify any vertical asymptotes, which occur where the function is undefined due to division by zero. 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:

Factor out the constant $4$ from the limit expression: $4\underset{x\to \mathrm{\infty }}{lim}\frac{x}{\sqrt{16x^2+5}}$.

Step 3.2:

Normalize the expression by dividing the numerator and the denominator by $x$: $4\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{x}{x}}{\sqrt{\frac{16x^2}{x^2}+\frac{5}{x^2}}}$.

Step 3.3:

Simplify the limit.

Step 3.3.1:

Cancel out $x$ in the numerator: $4\underset{x\to \mathrm{\infty }}{lim}\frac{1}{\sqrt{\frac{16x^2}{x^2}+\frac{5}{x^2}}}$.

Step 3.3.2:

Simplify the expression inside the square root.

Step 3.3.2.1:

Remove the common $x^2$ term: $4\underset{x\to \mathrm{\infty }}{lim}\frac{1}{\sqrt{16+\frac{5}{x^2}}}$.

Step 3.3.3:

Apply the limit rules to split the limit: $4\frac{\underset{x\to \mathrm{\infty }}{lim}1}{\underset{x\to \mathrm{\infty }}{lim}\sqrt{16+\frac{5}{x^2}}}$.

Step 3.3.4:

Evaluate the limit of the constant: $4\frac{1}{\underset{x\to \mathrm{\infty }}{lim}\sqrt{16+\frac{5}{x^2}}}$.

Step 3.3.5:

Apply the limit to the square root: $4\frac{1}{\sqrt{\underset{x\to \mathrm{\infty }}{lim}16+\frac{5}{x^2}}}$.

Step 3.3.6:

Split the limit inside the square root: $4\frac{1}{\sqrt{\underset{x\to \mathrm{\infty }}{lim}16+\underset{x\to \mathrm{\infty }}{lim}\frac{5}{x^2}}}$.

Step 3.3.7:

Evaluate the limit of the constant inside the square root: $4\frac{1}{\sqrt{16+\underset{x\to \mathrm{\infty }}{lim}\frac{5}{x^2}}}$.

Step 3.3.8:

Since the term $\frac{5}{x^2}$ approaches zero, the limit simplifies to $4\frac{1}{\sqrt{16}}$.

Step 3.4:

Simplify the expression to find the horizontal asymptote: $4\cdot \frac{1}{4}=1$.

Step 4:

Repeat the process for $x$ approaching negative infinity to find the horizontal asymptote on the left side.

Step 4.1:

Factor out the constant $4$ from the limit expression: $4\underset{x\to -\mathrm{\infty }}{lim}\frac{x}{\sqrt{16x^2+5}}$.

Step 4.2:

Normalize the expression by dividing the numerator and the denominator by $x$: $4\underset{x\to -\mathrm{\infty }}{lim}\frac{\frac{x}{x}}{-\sqrt{\frac{16x^2}{x^2}+\frac{5}{x^2}}}$.

Step 4.3:

Simplify the limit.

Step 4.3.1:

Cancel out $x$ in the numerator: $4\underset{x\to -\mathrm{\infty }}{lim}\frac{1}{-\sqrt{\frac{16x^2}{x^2}+\frac{5}{x^2}}}$.

Step 4.3.2:

Simplify the expression inside the square root.

Step 4.3.2.1:

Remove the common $x^2$ term: $4\underset{x\to -\mathrm{\infty }}{lim}\frac{1}{-\sqrt{16+\frac{5}{x^2}}}$.

Step 4.4:

Simplify the expression to find the horizontal asymptote: $4\cdot \frac{-1}{4}=-1$.

Step 5:

List the horizontal asymptotes: $y=1$ and $y=-1$.

Step 6:

Check for oblique asymptotes by attempting polynomial division. Since the function involves a square root, no oblique asymptotes can be found.

Step 7:

Compile the set of all asymptotes: No vertical asymptotes, horizontal asymptotes at $y=1$ and $y=-1$, and no oblique asymptotes.

Knowledge Notes:

To find the asymptotes of a function, you typically need to consider three types: vertical, horizontal, and oblique (slant) asymptotes.

1. Vertical Asymptotes: These occur where the function goes to infinity, typically at points where the denominator is zero. However, if the numerator and denominator both go to zero, you may need to use limits or simplify the expression further to determine if a vertical asymptote exists.

2. Horizontal Asymptotes: These are found by evaluating the limit of the function as $x$ approaches infinity or negative infinity. If the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator, the horizontal asymptote is $y=0$. If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients.

3. Oblique Asymptotes: If the degree of the numerator is exactly one more than the degree of the denominator, the function may have an oblique asymptote. You can find it by performing polynomial long division or synthetic division.

In the given problem, the function has no vertical asymptotes since the square root in the denominator is always defined for real numbers. The horizontal asymptotes are found by evaluating limits as $x$ approaches positive and negative infinity, which yield $y=1$ and $y=-1$, respectively. There are no oblique asymptotes because the function does not have a higher degree in the numerator compared to the denominator and also because of the presence of the square root.