Question

evaluate \\( \lim_{x \to \infty} f(x) \\) and \\( \lim_{x \to -\infty} f(x) \\) for the rational function. then give the horizontal asymptote of \\( f \\) (if any)
\\( f(x) = \frac{\sqrt{49x^2 + 10}}{6x + 7} \\)

find \\( \lim_{x \to -\infty} f(x) \\). select the correct choice, and, if necessary, fill in the answer box to complete your choice.
\\( \bigcirc \\) a. \\( \lim_{x \to -\infty} \frac{\sqrt{49x^2 + 10}}{6x + 7} = -\frac{7}{6} \\) (simplify your answer.)
\\( \bigcirc \\) b. the limit does not exist and is neither \\( \infty \\) nor \\( -\infty \\)

identify the horizontal asymptote(s). select the correct choice, and, if necessary, fill in the answer box(es) to complete your choice.
\\( \bigcirc \\) a. the function has two horizontal asymptotes. the top asymptote is \\( \square \\) and the bottom asymptote is \\( \square \\) (type equations. use integers or fractions for any numbers in the equations.)
\\( \bigcirc \\) b. the function has one horizontal asymptote, \\( \square \\) (type an equation. use integers or fractions for any numbers in the equation.)
\\( \bigcirc \\) c. the function has no horizontal asymptotes

Explanation:

Step1: Analyze \( \lim_{x \to +\infty} f(x) \)

For \( x \to +\infty \), \( \sqrt{49x^2 + 10} \approx \sqrt{49x^2} = 7x \) (since \( x>0 \), \( \sqrt{x^2}=x \)). So \( f(x)=\frac{\sqrt{49x^2 + 10}}{6x + 7} \approx \frac{7x}{6x + 7} \). Divide numerator and denominator by \( x \): \( \frac{7}{6 + \frac{7}{x}} \). As \( x \to +\infty \), \( \frac{7}{x} \to 0 \), so \( \lim_{x \to +\infty} f(x)=\frac{7}{6} \).

Step2: Analyze \( \lim_{x \to -\infty} f(x) \)

For \( x \to -\infty \), \( \sqrt{49x^2 + 10} \approx \sqrt{49x^2} = -7x \) (since \( x<0 \), \( \sqrt{x^2}=-x \)). So \( f(x)=\frac{\sqrt{49x^2 + 10}}{6x + 7} \approx \frac{-7x}{6x + 7} \). Divide numerator and denominator by \( x \) (note \( x<0 \), so \( \frac{-7x}{x}=-7 \), \( \frac{6x}{x}=6 \), \( \frac{7}{x}\to 0 \)): \( \frac{-7}{6 + \frac{7}{x}} \). As \( x \to -\infty \), \( \frac{7}{x} \to 0 \), so \( \lim_{x \to -\infty} f(x)=-\frac{7}{6} \).

Step3: Determine Horizontal Asymptotes

A horizontal asymptote \( y = L \) exists if \( \lim_{x \to +\infty} f(x)=L \) or \( \lim_{x \to -\infty} f(x)=L \). Here, \( \lim_{x \to +\infty} f(x)=\frac{7}{6} \) and \( \lim_{x \to -\infty} f(x)=-\frac{7}{6} \), so there are two horizontal asymptotes: \( y = \frac{7}{6} \) (top) and \( y = -\frac{7}{6} \) (bottom).

Answer:

For \( \lim_{x \to +\infty} f(x) \), the value is \( \frac{7}{6} \). For horizontal asymptotes, Option A: The top asymptote is \( y = \frac{7}{6} \) and the bottom asymptote is \( y = -\frac{7}{6} \).