📝 Rational functions limits at infinity (15 MCQs)

📖 Explanation: Dividing numerator and denominator by the highest power of $x$ in the denominator ($x$) gives $\frac{3+5/x}{6-8/x}$. As $x\to+\infty$, the terms $5/x$ and $8/x$ vanish, leaving $\frac{3}{6}= \tfrac12$. However, because the denominator grows faster than the numerator, the overall fraction approaches zero.

📖 Explanation: Factor the highest power of $x$ from both numerator and denominator: $\frac{x^{3}(2-5/x^{2})}{x^{3}(1+4/x^{3})}$. The fractions $5/x^{2}$ and $4/x^{3}$ go to zero as $x\to-\infty$. The remaining ratio of leading coefficients is $2/1=2$. Since both leading terms are negative for large negative $x$, the sign stays positive, giving a limit of $2$.

📖 Explanation: Divide by the highest power in the denominator ($x^{6}$): $\frac{-7+3/x^{4}}{4-5/x^{5}}$. As $x\to-\infty$, the terms with $x$ in the denominator vanish, leaving $-7/4$. Because the degree is even, the sign of the leading coefficients determines the sign of the limit, so the limit is $-\tfrac{7}{4}$.

📖 Explanation: For each function, divide numerator and denominator by $x^{2}$. The expressions become $\frac{5+2/x^{2}}{1-3/x^{2}}$ and $\frac{5-7/x^{2}}{1+4/x^{2}}$. The terms involving $1/x^{2}$ vanish, leaving the ratio of leading coefficients $5/1=5$ for both functions.

📖 Explanation: When degrees are equal, the limit equals the ratio of leading coefficients. For (i) the ratio is $1/2$; for (iii) it is $(-6)/(-6)=1$; both are finite and non‑zero. In (ii) the denominator’s degree exceeds the numerator’s, so the limit is $0$. Hence (i) and (iii) satisfy the condition.

📖 Explanation: The principle states that as $|x|$ becomes large, the term with the greatest exponent outweighs all lower‑degree terms. Consequently, the polynomial’s behavior mirrors that of its leading term, making the highest‑degree term the dominant factor in determining limits at infinity.

📖 Explanation: Dividing numerator and denominator by the highest power of $x$ present in the denominator isolates the leading terms. The remaining fractions involving lower powers of $x$ approach zero, leaving only the ratio of the leading coefficients. This reasoning explains why the limit equals that ratio.

📖 Explanation: Both numerator and denominator have even degree (8). After dividing by $x^{8}$ the expression approaches $-4/2=-2$. Since the leading coefficient of the numerator is negative and the denominator’s is positive, the limit is a negative finite number.

📖 Explanation: Dividing by $x^{5}$ yields $\frac{7-4/x^{2}+2/x^{4}-9/x^{5}}{-3+6/x^{3}-1/x^{5}}$. As $x\to+\infty$, the terms containing $1/x$ disappear, leaving the ratio of the leading coefficients $7/(-3)=-7/3$.

📖 Explanation: As $x$ grows without bound, $x^{2}$ grows even faster because squaring a large positive number yields a larger positive number. Hence the limit diverges to positive infinity.

📖 Explanation: When $n$ is odd, $\lim_{x\to-\infty}x^{n}$ is $-\infty$. For $n=3$, the cubic function preserves the sign of $x$; thus as $x$ approaches $-\infty$, the value also approaches $-\infty$.

📖 Explanation: Dividing numerator and denominator by $x$ gives $\frac{a+b/x}{c+d/x}$. As $x\to+\infty$, the fractions $b/x$ and $d/x$ vanish, leaving $\frac{a}{c}$. Therefore the limit $k$ must equal the ratio of the leading coefficients $a/c$.

📖 Explanation: Divide by $x^{4}$: $\frac{1-2/x^{2}+1/x^{4}}{1+3/x^{2}-5/x^{4}}$. The terms containing $1/x^{2}$ and $1/x^{4}$ go to zero, leaving the ratio $1/1=1$. Both numerator and denominator have even degree, so the sign remains positive.

📖 Explanation: If a polynomial $p(x)$ has limit $L$ as $x\to\pm\infty$, then $-c\,p(x)$ (with $c>0$) has limit $-cL$. The negative constant flips the sign of the limit while preserving its magnitude, because limits respect scalar multiplication.

📖 Explanation: After division, the expression becomes a ratio of a polynomial whose numerator contains the leading term of the original numerator and a denominator whose leading term is $1$. All other terms contain negative powers of $x$ and therefore vanish as $|x|$ grows, leaving the limit determined solely by the leading coefficients.