Question

a. Evaluate \(\lim _{x \rightarrow \infty} f(x)\) and \(\lim _{x \rightarrow-\infty} f(x),\) and then identify any horizontal asymptotes. b. Find the vertical asymptotes. For each vertical asymptote \(x=a\), evaluate \(\lim _{x \rightarrow a^{-}} f(x)\) and \(\lim _{x \rightarrow a^{+}} f(x)\). $$f(x)=\frac{\sqrt{x^{2}+2 x+6}-3}{x-1}$$

Short answer

Answer: The limits of the function as \(x\) approaches infinity and negative infinity are both equal to \(\frac{1}{2}\). b. Identify any horizontal asymptotes. Answer: The horizontal asymptote is at \(y = \frac{1}{2}\). c. What is the vertical asymptote of the function? Answer: The vertical asymptote is at \(x = 1\). d. What are the limits of the function as \(x\) approaches the vertical asymptote from the left and right? Answer: The limits of the function as \(x\) approaches the vertical asymptote from the left (\(x \rightarrow 1^{-}\)) and right (\(x \rightarrow 1^{+}\)) are both equal to \(1\).

Step-by-step solution

a. Evaluating Limits as \(x\) Approaches Infinity and Negative Infinity

To simplify the function, we can use the conjugate of the numerator as follows: $$ f(x)=\frac{\sqrt{x^{2}+2 x+6}-3}{x-1} = \frac{(\sqrt{x^{2}+2 x+6}-3)((\sqrt{x^{2}+2 x+6}+3))}{(x-1)((\sqrt{x^{2}+2 x+6}+3))} $$ The numerator simplifies to: $$ (\sqrt{x^{2}+2 x+6}-3)(\sqrt{x^{2}+2 x+6}+3) = (x^2 + 2x + 6) - 9 = x^2 + 2x - 3 $$ So the simplified function is: $$ f(x) = \frac{x^{2}+2 x-3}{(x-1)(\sqrt{x^{2}+2 x+6}+3)} $$ Now, let's evaluate the limits as \(x\) approaches infinity and negative infinity: $$ \lim_{x\rightarrow \infty}\frac{x^{2}+2 x-3}{(x-1)(\sqrt{x^{2}+2 x+6}+3)} = \lim_{x\rightarrow \infty}\frac{x^2(1+\frac{2}{x}-\frac{3}{x^2})}{x^2(1-\frac{1}{x})(1+\frac{2}{x}+\sqrt{1+\frac{2}{x}+\frac{6}{x^2}})} $$ As \(x\) approaches infinity, the terms with \(x\) in the denominator will approach zero: $$ \lim_{x\rightarrow \infty}\frac{1+\frac{2}{x}-\frac{3}{x^2}}{(1-\frac{1}{x})(1+\frac{2}{x}+\sqrt{1+\frac{2}{x}+\frac{6}{x^2}})} = \frac{1}{(1)(1+\sqrt{1})} = \frac{1}{2} $$ The limit as \(x\) approaches negative infinity is the same, since the function is an even function: $$ \lim_{x\rightarrow -\infty}f(x) = \lim_{x\rightarrow \infty}f(x) = \frac{1}{2} $$ Therefore, the horizontal asymptote is \(y=\frac{1}{2}\).

b. Finding the Vertical Asymptotes and Evaluating Limits as \(x \rightarrow a\) from the Left and Right

To find the vertical asymptote(s), we need to determine the values of \(x\) that make the denominator of the function zero: $$ (x-1)(\sqrt{x^{2}+2 x+6}+3) = 0 $$ It's easy to see that \(x = 1\) makes the first term in the denominator zero. So the vertical asymptote is \(x = 1\). Now, we will evaluate the limits as \(x\) approaches \(1\) from the left and the right: $$ \lim_{x\rightarrow 1^{-}}\frac{x^{2}+2 x-3}{(x-1)(\sqrt{x^{2}+2 x+6}+3)} \quad \text{and} \quad \lim_{x\rightarrow 1^{+}}\frac{x^{2}+2 x-3}{(x-1)(\sqrt{x^{2}+2 x+6}+3)} $$ For both limits, we can use L'Hôpital's Rule by taking the derivative of the numerator and the denominator: $$ \frac{d}{dx}(x^2+2x-3) = 2x+2 \\ \frac{d}{dx}((x-1)(\sqrt{x^2+2x+6}+3)) = (1)(\sqrt{x^{2}+2 x+6}+3) + (x-1)\frac{1}{2}(\frac{(2x+2)(2)}{\sqrt{x^2+2x+6}}) $$ Now let's evaluate both left and right limits: $$ \lim_{x\rightarrow 1^{-}}\frac{(2x+2)}{(\sqrt{x^{2}+2 x+6}+3) + (x-1)\frac{1}{2}(\frac{(2x+2)(2)}{\sqrt{x^2+2x+6}})} = \lim_{x\rightarrow 1^{-}}\frac{2(x+1)}{(\sqrt{x^{2}+2 x+6}+3) + \frac{x-1}{\sqrt{x^2+2x+6}}(2x+2)} \quad (*) \\ \lim_{x\rightarrow 1^{+}}\frac{(2x+2)}{(\sqrt{x^{2}+2 x+6}+3) + (x-1)\frac{1}{2}(\frac{(2x+2)(2)}{\sqrt{x^2+2x+6}})} = \lim_{x\rightarrow 1^{+}}\frac{2(x+1)}{(\sqrt{x^{2}+2 x+6}+3) + \frac{x-1}{\sqrt{x^2+2x+6}}(2x+2)} \quad (**) $$ By plugging \(x=1\) into the limit expressions above, we get: $$ (*) = \frac{2(1+1)}{(\sqrt{1^2+2*1+6}+3) + \frac{1-1}{\sqrt{1^2+2*1+6}}(2*1+2)} = \frac{4}{(3+3)} = 1 \\ (**) = 1 \\ $$ Thus, we have: $$ \lim_{x\rightarrow 1^{-}}f(x) = 1 \quad \text{and} \quad \lim_{x\rightarrow 1^{+}}f(x) = 1 $$

Key concepts

Horizontal Asymptotes

When analyzing rational functions, horizontal asymptotes tell us the behavior of the function as it moves toward positive or negative infinity. For a function like the one in our exercise, which simplifies during evaluation, horizontal asymptotes represent the values that the function approaches as \( x \) goes to very large or very small values.
In this exercise, we tackled the limits at infinity. After simplifying the given function \( f(x) \), we evaluated the expression's behavior when \( x \) goes to infinity and negative infinity. Both limits turned out to be \( \frac{1}{2} \).
This tells us that the output of the function approaches the constant value \( y = \frac{1}{2} \) regardless of whether \( x \) is moving positively or negatively infinitely far. Therefore, we conclude that the function has a horizontal asymptote at \( y = \frac{1}{2} \). This means that the graph of the function will never intersect this line as \( x \) becomes extremely large or small.

Vertical Asymptotes

Vertical asymptotes are lines that the graph of a function approaches but never touches as the input \( x \) approaches a certain value. These typically occur when the denominator of a rational function becomes zero while the numerator doesn't.
In our step-by-step solution, we determined the vertical asymptote by looking for values of \( x \) that cause the denominator of the function to be zero. Solving \((x-1)(\sqrt{x^2+2x+6}+3) = 0\) revealed that \( x = 1 \) is where the denominator becomes zero. As a result, \( x = 1 \) is a vertical asymptote.
This vertical asymptote at \( x = 1 \) represents a boundary line where the function becomes undefined, indicating that the graph of \( f(x) \) will approach but not intersect this line.

Limits at Infinity

Limits at infinity help us understand how a function behaves as \( x \) approaches the extreme ends of the number line (both positive and negative). They are essential for identifying horizontal asymptotes.
To evaluate the limit at infinity for our function, we first simplified it using algebraic manipulation, including rationalizing the numerator. After simplifying, we looked at the behavior of the expression as \( x \to \infty \) and \( x \to -\infty \).
Since terms with \( x \) in the denominator become negligible as \( x \) turns very large (approaching zero), this allowed us to find the limits quickly. The main takeaway was understanding that these limits reveal key asymptotic behavior of functions as \( x \) trends towards large magnitudes.

L'Hôpital's Rule

L'Hôpital's Rule is a helpful mathematical technique used to evaluate limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When standard algebraic techniques are difficult, L'Hôpital's Rule becomes effective.
In our exercise, we applied L'Hôpital's Rule to analyze the limits around \( x = 1 \), where potential vertical asymptotes are located. By differentiating the numerator and the denominator of the step involving this limit, we bypass tricky algebra and focus on simpler derivatives. This method confirmed consistent results of both \( \lim_{x\to1^-} f(x) \) and \( \lim_{x\to1^+} f(x) \) leading to 1.
Using this rule ensures precision when solving complex limits, particularly useful when encountering indeterminate cases during examinations of rational functions.