Question

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

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

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

Explanation:

For $\boldsymbol{\lim_{x \to -\infty} f(x)}$:

Step1: Simplify the square root

For \( x \to -\infty \), \( \sqrt{49x^2 + 10} = |x|\sqrt{49 + \frac{10}{x^2}} \). Since \( x \to -\infty \), \( |x| = -x \). So we have \( \sqrt{49x^2 + 10} = -x\sqrt{49 + \frac{10}{x^2}} \).

Step2: Substitute into the function

Substitute into \( f(x) = \frac{\sqrt{49x^2 + 10}}{6x + 7} \), we get \( \frac{-x\sqrt{49 + \frac{10}{x^2}}}{6x + 7} \).

Step3: Divide numerator and denominator by \( x \)

Divide numerator and denominator by \( x \) (note \( x < 0 \), so dividing by \( x \) is like dividing by a negative number, but we just factor out \( x \) from denominator: \( 6x + 7 = x(6 + \frac{7}{x}) \)). So the function becomes \( \frac{-x\sqrt{49 + \frac{10}{x^2}}}{x(6 + \frac{7}{x})} \). Cancel out \( x \) (remember \( x
eq 0 \) as \( x \to -\infty \)): \( \frac{-\sqrt{49 + \frac{10}{x^2}}}{6 + \frac{7}{x}} \).

Step4: Take the limit as \( x \to -\infty \)

As \( x \to -\infty \), \( \frac{10}{x^2} \to 0 \) and \( \frac{7}{x} \to 0 \). So we have \( \lim_{x \to -\infty} \frac{-\sqrt{49 + 0}}{6 + 0} = \frac{-7}{6} \).

Answer:

For \( \lim_{x \to \infty} f(x) \), the answer is \( \frac{7}{6} \) (Option A). For \( \lim_{x \to -\infty} f(x) \), the limit is \( \frac{-7}{6} \) (Option A with value \( -\frac{7}{6} \)). The horizontal asymptote is \( y = \frac{7}{6} \) (since \( \lim_{x \to \infty} f(x) = \lim_{x \to -\infty} f(x) \) is not true here? Wait, no, wait: Wait, when \( x \to \infty \), we had \( \frac{7}{6} \), when \( x \to -\infty \), we have \( -\frac{7}{6} \)? Wait, no, let's re - check. Wait, when \( x \to -\infty \), \( |x|=-x \), so \( \sqrt{49x^2 + 10}=|x|\sqrt{49+\frac{10}{x^2}}=-x\sqrt{49 + \frac{10}{x^2}} \) (because \( x \) is negative, \( |x|=-x \)). Then \( f(x)=\frac{-x\sqrt{49+\frac{10}{x^2}}}{6x + 7} \). Now, divide numerator and denominator by \( x \) ( \( x<0 \) ), so \( \frac{-x\sqrt{49+\frac{10}{x^2}}}{x(6+\frac{7}{x})}=\frac{-\sqrt{49+\frac{10}{x^2}}}{6+\frac{7}{x}} \). Now, as \( x\to-\infty \), \( \frac{10}{x^2}\to0 \) and \( \frac{7}{x}\to0 \), so the limit is \( \frac{-\sqrt{49}}{6}=-\frac{7}{6} \). And for horizontal asymptote, since \( \lim_{x \to \infty}f(x)=\frac{7}{6} \) and \( \lim_{x \to -\infty}f(x)=-\frac{7}{6} \), wait, no, that can't be. Wait, maybe I made a mistake. Wait, the function is \( f(x)=\frac{\sqrt{49x^2 + 10}}{6x + 7} \). Let's factor out \( x^2 \) from the square root: \( \sqrt{49x^2+10}=\sqrt{x^2(49 + \frac{10}{x^2})}=|x|\sqrt{49+\frac{10}{x^2}} \). When \( x\to\infty \), \( |x| = x \), so \( \frac{x\sqrt{49+\frac{10}{x^2}}}{x(6+\frac{7}{x})}=\frac{\sqrt{49+\frac{10}{x^2}}}{6+\frac{7}{x}}\to\frac{7}{6} \). When \( x\to-\infty \), \( |x|=-x \), so \( \frac{-x\sqrt{49+\frac{10}{x^2}}}{x(6+\frac{7}{x})}=\frac{-\sqrt{49+\frac{10}{x^2}}}{6+\frac{7}{x}}\to\frac{-7}{6} \). But for horizontal asymptote, a horizontal asymptote \( y = L \) exists if \( \lim_{x \to \infty}f(x)=L \) or \( \lim_{x \to -\infty}f(x)=L \). But in this case, since the two limits are different, wait, no, wait the problem says "give the horizontal asymptote of \( f \) (if any)". Wait, actually, when the degrees of the numerator and denominator (after considering the square root, the numerator is degree 1, denominator is degree 1) are equal, we look at the leading coefficients. But when \( x\to\infty \), the leading coefficient ratio is \( \frac{7}{6} \), when \( x\to-\infty \), it's \( \frac{-7}{6} \). But actually, the horizontal asymptote is determined by the limit as \( x\to\infty \) and \( x\to-\infty \). If both limits are equal, that's the horizontal asymptote. If they are different, there is no horizontal asymptote? Wait, no, let's check the definition. The horizontal asymptote of a function \( y = f(x) \) is a horizontal line \( y = L \) such that \( \lim_{x \to \infty}f(x)=L \) or \( \lim_{x \to -\infty}f(x)=L \). But in some cases, if \( \lim_{x \to \infty}f(x)=L_1 \) and \( \lim_{x \to -\infty}f(x)=L_2 \) with \( L_1
eq L_2 \), then there are two horizontal asymptotes? Wait, no, the standard definition: A horizontal asymptote is a line \( y = L \) where \( \lim_{x \to \infty}f(x)=L \) or \( \lim_{x \to -\infty}f(x)=L \). So in this case, as \( x\to\infty \), the limit is \( \frac{7}{6} \), so \( y=\frac{7}{6} \) is a horizontal asymptote, and as \( x\to-\infty \), the limit is \( -\frac{7}{6} \), so \( y = -\frac{7}{6} \) is also a horizontal asymptote? Wait, but let's check the function again. The function is \( f(x)=\frac{\sqrt{49x^2 + 10}}{6x + 7} \). Let's rewrite it for \( x<0 \): let \( t=-x \), so \( t\to\infty \) as \( x\to-\infty \). Then \( f(x)=\frac{\sqrt{49t^2 + 10}}{-6t + 7}=\frac{t\sqrt{49+\frac{10}{t^2}}}{t(-6+\frac{7}{t})}=\frac{\sqrt{49+\frac{10}{t^2}}}{-6+\frac{7}{t}}\to\frac{7}{-6}=-\frac{7}{6} \) as \( t\to\infty \) (i.e., \( x\to-\infty \)). So for \( \lim_{x \to \infty}f(x) \), the answer is \( \frac{7}{6} \) (Option A). For \( \lim_{x \to -\infty}f(x) \), the limit is \( -\frac{7}{6} \) (so Option A with value \( -\frac{7}{6} \)). And the horizontal asymptotes are \( y = \frac{7}{6} \) (from \( x\to\infty \)) and \( y=-\frac{7}{6} \) (from \( x\to-\infty \))? Wait, but the problem says "give the horizontal asymptote of \( f \) (if any)". Maybe the problem considers that when the degrees are equal, we take the ratio of leading coefficients, but we have to consider the sign of \( x \) in the square root. So, to summarize:

• \( \lim_{x \to \infty} f(x)=\frac{7}{6} \) (Option A)
• \( \lim_{x \to -\infty} f(x)=-\frac{7}{6} \) (Option A, fill in \( -\frac{7}{6} \))
• Horizontal asymptotes: \( y = \frac{7}{6} \) and \( y=-\frac{7}{6} \) (but if we consider the standard case where we look at \( x\to\infty \) first, \( y = \frac{7}{6} \) is a horizontal asymptote, and \( y=-\frac{7}{6} \) is also a horizontal asymptote)