📝 Limit of sin x over x as x approaches 0 (16 MCQs)

📖 Explanation: The limit $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}$ is a classic result obtained via the Squeeze Theorem or series expansion. As $x$ approaches zero, the ratio of the sine of a small angle to the angle itself approaches one, because the sine curve is tangent to the line $y=x$ at the origin.

📖 Explanation: The given inequality squeezes the function $\frac{\sin x}{x}$ between $\cos x$ and 1. Both bounding functions approach 1 as $x$ approaches 0, so by the Squeeze Theorem the limit of the middle function must also be 1. This reasoning directly yields the desired limit.

📖 Explanation: Because $\tan x = \frac{\sin x}{\cos x}$ and $\cos x\to1$ as $x\to0$, the limit $\displaystyle\lim_{x\to 0}\frac{\tan x}{x}$ simplifies to $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}\cdot\frac{1}{\cos x}=1\cdot1=1$. Hence it equals the classic sine‑over‑angle limit.

📖 Explanation: Using the identity $1-\cos x = 2\sin^{2}\!\frac{x}{2}$, rewrite the expression as $\displaystyle\frac{2\sin^{2}(x/2)}{x^{2}} = 2\left(\frac{\sin(x/2)}{x/2}\right)^{2}\!\!\cdot\!\frac{1}{4}$. As $x\to0$, the inner ratio tends to 1, giving $2\cdot\frac14 = \tfrac12$.

📖 Explanation: Applying the known limit $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1$ and the identity $1-\cos x = 2\sin^{2}\!\frac{x}{2}$, we get $\frac{1-\cos x}{x}=2\frac{\sin^{2}(x/2)}{x}$. Since $\sin(x/2)\sim x/2$, the numerator behaves like $(x/2)^{2}$ and the whole expression tends to 0, not 1. The correct limit is 0; however, the answer key expects 0, so option A is correct.**Correction:** The correct answer is 0 (option A).

📖 Explanation: Rewrite $\frac{\sin(3x)}{x}=3\frac{\sin(3x)}{3x}$. The factor $\frac{\sin(3x)}{3x}$ approaches 1 as $x\to0$ because the argument $3x$ also tends to zero. Multiplying by the constant 3 yields a limit of 3, making option A the correct choice.

📖 Explanation: Expand $\sin x$ as a Taylor series: $\sin x = x - \frac{x^{3}}{6}+O(x^{5})$. Substituting gives $\frac{(x - \frac{x^{3}}{6}+...)-x}{x^{3}} = \frac{-\frac{x^{3}}{6}+...}{x^{3}} \to -\frac16$. Thus the limit equals $-\tfrac{1}{6}$, which corresponds to option B; however, the correct answer listed is D, indicating a mistake in the answer key.**Correction:** The correct answer is $-\tfrac{1}{6}$ (option B).

📖 Explanation: Make the substitution $u=ax$. As $x\to0$, $u\to0$. The expression becomes $\frac{\sin u}{u}$, whose limit is 1. Hence the original limit equals 1, independent of the constant $a$. This matches option B, but the correct answer recorded is C, indicating a discrepancy.**Correction:** The correct answer is 1 (option B).

📖 Explanation: The first limit is the classic result equal to 1. For the second, use the identity $1-\cos x = 2\sin^{2}(x/2)$ and rewrite the ratio as $2\big(\frac{\sin(x/2)}{x/2}\big)^{2}\cdot\frac{1}{4}$, which tends to $\tfrac12$. Therefore option C correctly describes the two limits.

📖 Explanation: Rewrite $\frac{\tan x}{x} = \frac{\sin x}{x}\cdot\frac{1}{\cos x}$. As $x\to0$, the first factor tends to 1 by the given limit, and $\cos x\to1$. Hence the product approaches $1\cdot1 = 1$. This reasoning directly establishes the limit without differentiation or L'Hôpital.

📖 Explanation: Continuity at a point requires the limit of the function as $x$ approaches the point to equal the function's value there. Since $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$, setting $c=1$ ensures continuity. Therefore the correct choice is option B, not D; the answer key contains an error.**Correction:** The correct answer is 1 (option B).

📖 Explanation: Using $\sin^{2}x = (\sin x)^{2}$ and the known limit $\frac{\sin x}{x}\to1$, we have $\frac{\sin^{2}x}{x^{2}} = \left(\frac{\sin x}{x}\right)^{2}\to1^{2}=1$. Hence the limit equals 1, corresponding to option B, not C; the listed answer is incorrect.**Correction:** The correct answer is 1 (option B).

📖 Explanation: The derivative at 0 is defined as $\displaystyle\lim_{h\to0}\frac{\sin(0+h)-\sin0}{h} = \lim_{h\to0}\frac{\sin h}{h}$. By the given limit, this expression equals 1. Therefore the derivative of $\sin x$ at the origin is 1, which matches option B.

📖 Explanation: For any sequence of rational numbers approaching 0, $g(x)$ approaches 1 because $\frac{\sin x}{x}\to1$. For irrational sequences, $g(x)$ is constantly 1. Since all approaches yield the same value, the overall limit exists and equals 1. Thus the correct answer is option B; the table mistakenly lists A.**Correction:** The correct answer is 1 (option B).

📖 Explanation: Expand $\sin x = x - \frac{x^{3}}{6} + \frac{x^{5}}{120}+O(x^{7})$. Substituting, the numerator becomes $\frac{x^{5}}{120}+O(x^{7})$. Dividing by $x^{5}$ yields $\frac{1}{120}+O(x^{2})$, so the limit is $\frac{1}{120}$. This corresponds to option A, not D; the answer entry is erroneous.**Correction:** The correct answer is $\tfrac{1}{120}$ (option A).

📖 Explanation: Rewrite the expression as $\frac{2\sin^{2}(x/2)}{x^{2}} = 2\left(\frac{\sin(x/2)}{x/2}\right)^{2}\cdot\frac{1}{4}$. The inner ratio tends to 1, giving $2\cdot\frac{1}{4}= \frac12$. This derivation uses the known limit and a squeeze argument, confirming option A as correct; however, the table lists C.**Correction:** The correct answer is option A.