📝 Limit laws sum product quotient (20 MCQs)

📖 Explanation: Applying the constant multiple law gives $\lim 2f(x)=2\cdot3=6$. The difference law yields $\lim (2f(x)-g(x))=6-(-2)=8$. Therefore the limit equals 8, making option A correct.

📖 Explanation: The power law for limits at infinity states that the limit of a power equals the power of the limit, provided the original limit exists. Hence $\lim (h(x))^{3}=L^{3}$, which corresponds to option B.

📖 Explanation: The given information only tells us that the magnitude of $f(x)$ shrinks to zero while its reciprocal grows without bound. It provides no insight into the sign of $f(x)$ as $x\to+\infty$; thus we cannot conclude which side it approaches from, making D correct.

📖 Explanation: The factor 5 can be pulled out of the limit, giving $5\cdot\lim_{x\to-\infty}\frac{1}{x}$. Since $\frac{1}{x}$ tends to 0 as $x\to-\infty$, the product is $5\cdot0=0$. Hence option A is correct.

📖 Explanation: When the numerator approaches a finite number (4) and the denominator grows without bound, the quotient approaches 0. This follows from the quotient law for limits at infinity, so the correct answer is 0, listed as option B.

📖 Explanation: Divide numerator and denominator by $x^{2}$ to obtain $\lim_{x\to+\infty}(1-\frac{1}{x})$. The term $\frac{1}{x}$ tends to 0, leaving a limit of 1. Since option D corresponds to 1, it is the correct choice.

📖 Explanation: Both $\frac{1}{x}$ and $\frac{1}{x^{2}}$ shrink to zero as $x$ grows without bound. The limit laws for powers of $x$ confirm that each limit is 0, making option A correct.

📖 Explanation: Rewrite the expression as $\left[\left(1+\frac{1}{2x}\right)^{2x}\right]$ and recognize the familiar form $(1+\frac{1}{n})^{n}$ with $n=2x$. The known limit $\lim_{n\to\infty}(1+\frac{1}{n})^{n}=e$ directly yields the result, so option D is correct.

📖 Explanation: Apply the constant multiple law to write $\frac{k}{f(x)}=k\cdot\frac{1}{f(x)}$. The reciprocal limit law gives $\lim \frac{1}{f(x)}=\frac{1}{L}$. Multiplying by $k$ yields $\frac{k}{L}$. Since this matches option C, it is correct.

📖 Explanation: The intermediate value property is a theorem about continuous functions on closed intervals, not a limit law. All other listed laws (sum, product, quotient) extend to limits at infinity, so option D is the correct answer.

📖 Explanation: When $n$ is even, $\frac{1}{x^{n}}$ is positive for both large positive and negative $x$, and each tends to 0. When $n$ is odd, the sign changes but the magnitude still approaches 0. Thus both limits are 0, making option B the correct choice.

📖 Explanation: The product law gives $\lim f(x)g(x)=ab$. Adding a constant uses the sum law, so $\lim (f(x)g(x)+3)=ab+3$. This matches option A.

📖 Explanation: Separate the fraction: $\frac{3x^{2}}{x^{2}}-\frac{2x}{x^{2}}=3-\frac{2}{x}$. The term $\frac{2}{x}$ tends to 0 as $x\to+\infty$, leaving a limit of 3. Hence option A is correct.

📖 Explanation: The expression is precisely the classic sequence defining $e$. By recognizing $(1+\frac{1}{x})^{x}$ as $(1+\frac{1}{n})^{n}$ with $n=x$ and applying the known limit $\lim_{n\to\infty}(1+\frac{1}{n})^{n}=e$, we justify the result, so option D is correct.

📖 Explanation: The expression $0^{\infty}$ is an indeterminate form; the limit can depend on the rates at which $f(x)$ approaches zero and $g(x)$ diverges. Limit laws do not resolve this case, so the appropriate answer is that the form is indeterminate, option D.

📖 Explanation: A constant function does not change with $x$. The limit law for constants at infinity states that the limit equals the constant itself, so $\lim_{x\to-\infty}7=7$. This corresponds to option B.

📖 Explanation: The power law extends to fractional exponents: $\lim (f(x))^{1/2}=(\lim f(x))^{1/2}$ when the limit exists and is non‑negative. Since $L>0$, we may apply this law directly, giving option A.

📖 Explanation: For each fixed $n$, $\lim_{x\to+\infty}\frac{x}{n}=+\infty$; then $\lim_{n\to\infty}(+\infty)$ is undefined. Conversely, for each fixed $x$, $\lim_{n\to\infty}\frac{x}{n}=0$; then $\lim_{x\to+\infty}0=0$. Thus the first limit does not exist while the second equals 0, matching option C.

📖 Explanation: The passage explicitly mentions that the limit laws in Theorem 1.2.2 carry over unchanged to limits at $+\infty$ and $-\infty$. Therefore Theorem 1.2.2 is the correct reference, option A.

📖 Explanation: The constant function limit law asserts that a constant remains unchanged under a limit, even at infinity. Hence $\lim_{x\to+\infty}k=k$. This statement corresponds to option C.