Quick reasoning: \[\lim_{x \to \infty}\left( x – x^2 \right) = \lim_{x \to \infty}\left(- x^2 \right) = -\infty \quad \cmark\]

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\[ \begin{align*} \lim_{x \to \infty}\frac{4x^3 + 2x -24}{x^4 – x^2 + 84 } &= \lim_{x \to \infty}\frac{4x^3}{x^4} \\[8px] &= 0 \quad \cmark \end{align*} \] because the highest power in the denominator is greater than the highest power in the numerator. (That’s it; the reasoning is that simple.)

Conceptually, the growth in the denominator “wins out” over the growth in the numerator, meaning the denominator grows large at a faster rate than the numerator does, and so the fraction tends toward zero as x grows and Grows and GROWS in the positive direction.

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\begin{align*} \lim_{x \to \infty} \frac{x + 7}{x^3 -x +2} &= \lim_{x \to \infty} \frac{x}{x^3} \\[8px] &= 0 \implies \quad\text{ (B)} \quad \cmark \end{align*} because the highest power in the denominator is greater than the highest power in the numerator. (That’s it; the reasoning is that simple.)

Conceptually, the growth in the denominator “wins out” over the growth in the numerator, meaning the denominator grows large at a faster rate than the numerator does, and so the fraction tends toward zero as x grows and Grows and GROWS in the positive direction.

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We know immediately that the limit does not exist (DNE), because the highest power in the numerator $\left(x^3 \right)$ is larger than the highest power in the denominator $\left( x^2\right)$. Conceptually, the numerator “wins” over the denominator as x grows.

More specifically, we know that the limit is either $\infty$ or $-\infty$. To determine which, we use our usual approach and look at just the term with the highest power in the numerator and the term with the highest power in the denominator: \[ \begin{align*} \lim_{x \to \infty}\frac{x^3 +2}{3x^2 + 4} &= \lim_{x \to \infty}\frac{x^3}{3x^2} \\[8px] &= \lim_{x \to \infty}\frac{x}{3} \\[8px] &= \infty \quad \cmark \end{align*} \] The second line in the solution shows that the function approaches $\dfrac{x}{3}$  as x grows large, matching what the graph shows. The limit at infinity is (positive) $\infty$ because the function grows in the positive y-direction forever as x grows larger and Larger in the positive direction.

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We know immediately that the limit does not exist (DNE), because the highest power in the numerator $\left(x^2 \right)$ is larger than the highest power in the denominator $\left( x\right)$. Conceptually, the numerator “wins” over the denominator as x grows. \[\lim_{x \to \infty}\frac{x^2 + 3x}{x+1} = \lim_{x \to \infty}\frac{x^2}{x} = \infty \implies \quad\text{ (C)} \quad \cmark\] To show that this is the case more formally, divide each term in the fraction by the largest power that appears in the denominator, $x$: \begin{align*} \lim_{x \to \infty} \frac{x^2 + 3x}{x+1} &= \lim_{x \to \infty} \frac{\displaystyle{\frac{x^2}{x} + \frac{3x}{x}}}{\displaystyle{\frac{x}{x}+\frac{1}{x}}}\\[12px] &= \lim_{x \to \infty} \frac{x +3}{1 + \frac{1}{x}} \\[12px] \end{align*} Now $\displaystyle{\lim_{x \to \infty} \frac{1}{x} = 0}$, and hence the limit of the denominator is $\displaystyle{\lim_{x \to \infty}(1+ \frac{1}{x}) = 1}$. The numerator, on the other hand, grows without bound: $\displaystyle{\lim_{x \to \infty}(x +3) = \infty}$. Hence \begin{align*} \phantom{\lim_{x \to \infty} \frac{x^2 + 3x}{x+1}} & \\ \lim_{x \to \infty} \frac{x +3}{1 + \cancelto{0}{\frac{1}{x}}} &= \infty \implies \quad\text{ (C)} \quad \cmark \end{align*}

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To determine the limit at infinity we need only look at the term with the highest power in the numerator, and the term with the highest power in the denominator. In this problem, those powers are the same: $x^2$. The answer is then the ratio of the coefficients of those terms: \[ \begin{align*} \lim_{x \to \infty}\frac{5x^2 -7}{3x^2 + 8} &= \lim_{x \to \infty}\frac{5x^2}{3x^2} \\[8px] &= \frac{5}{3} \quad \cmark \end{align*} \] Conceptually, the numerator and denominator are growing at the same rate, as modified only by the coefficients of those largest terms.

By the way, the graph shows that the line $y = \dfrac{5}{3}$  is a horizontal asymptote for this function: the function’s curve gets arbitrarily close to that line as $x \to \infty$.

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Asking for the horizontal asymptotes is simply another way of asking what the function does as $x \to \infty$ and $x \to -\infty$. Your intuition is probably that as $x$ gets very large (in the positive or negative direction), the $x^2$ terms in the numerator and denominator dominate, and the function becomes essentially $\displaystyle{\frac{5x^2}{3x^2}}$, and hence the asymptote is $y = \dfrac{5}{3}$. This is correct.

To show that this is the case, divide each term in the fraction by the largest power in the denominator, $x^2$:
\begin{align*} \lim_{x \to \infty} \frac{5x^2 + x -3}{3x^2 – 2x + 5} &= \lim_{x \to \infty} \frac{\displaystyle{\frac{5x^2}{x^2} + \frac{x}{x^2} – \frac{3}{x^2}}}{\displaystyle{\frac{3x^2}{x^2} -\frac{2x}{x^2} + \frac{5}{x}}}\\ \\ &= \lim_{x \to \infty} \frac{\displaystyle{5 + \frac{1}{x} – \frac{3}{x^2}}}{\displaystyle{3 -\frac{2}{x} + \frac{5}{x}}}\\ \\ &= \frac{5}{3} \quad \cmark \end{align*} Similarly, \begin{align*} \lim_{x \to -\infty} \frac{5x^2 + x -3}{3x^2 – 2x + 5} &= \lim_{x \to \infty} \frac{\displaystyle{5 + \frac{1}{x} – \frac{3}{x^2}}}{\displaystyle{3 -\frac{2}{x} + \frac{5}{x}}}\\ &= \frac{5}{3} \quad \cmark \end{align*} Hence the horizontal asymptote is $y = 5/3$. See the graph.

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