📝 Trigonometric functions continuity (15 MCQs)

📖 Explanation: Because both sine and cosine are continuous, the limit of their quotient exists when the denominator is non‑zero. Hence $\lim_{x\to c}\tan x = \lim_{x\to c}\frac{\sin x}{\cos x}= \frac{\sin c}{\cos c}$. This matches option A.

📖 Explanation: The sine function is defined and continuous for every real number, while $\sec x = 1/\cos x$ is undefined wherever $\cos x = 0$. At $x = \pi/2$ the cosine equals zero, so $\sec x$ is not defined and thus not continuous. Only $\sin x$ remains continuous, giving option B.

📖 Explanation: Using the standard limit $\lim_{x\to 0}\frac{\sin(kx)}{x}=k$ for any constant $k$, we substitute $k=5$. The limit evaluates to 5, which corresponds to option C.

📖 Explanation: The function equals $\sin x$ everywhere except at $x=\pi$ where it is artificially set to 2. Since $\lim_{x\to \pi}\sin x = 0\neq 2$, the limit and the function value differ, so the function is not continuous at $x=\pi$. Option B states this reason.

📖 Explanation: At $x = \pi/2$ the cosine term in $\tan x = \sin x / \cos x$ approaches zero, causing the quotient to blow up. The limit does not exist (it tends to $\pm\infty$), so $\tan x$ is discontinuous there. Option B captures this.

📖 Explanation: The step $\lim_{x\to c}\cos(g(x)) = \cos(\lim_{x\to c}g(x))$ relies on the fact that cosine is a continuous function on its entire domain. This is precisely the continuity property of cosine, making option C correct.

📖 Explanation: Continuity alone guarantees that $\sin x$ does not exceed its known bounds of $[-1,1]$ on the whole real line, so it is bounded. The other statements require additional properties such as differentiability or invertibility, which are not implied by continuity. Hence option C is correct.

📖 Explanation: The limit $\lim_{x\to 0}\frac{\sin x}{x}=1$ is a classic result. Defining $h(0)=1$ makes the function value match the limit, ensuring continuity at $x=0$. Therefore the function is continuous, corresponding to option A.

📖 Explanation: If $g$ is continuous at $c$ then $\lim_{x\to c}g(x)=g(c)$. If $f$ is continuous at $g(c)$ then $\lim_{y\to g(c)}f(y)=f(g(c))$. Combining the two gives $\lim_{x\to c}f(g(x)) = f(g(c))$. This is exactly option A.

📖 Explanation: The secant function is defined as $\sec x = 1/\cos x$. It is undefined wherever $\cos x = 0$, which occurs at odd multiples of $\pi/2$. Hence its natural domain is $\mathbb{R}\\{(2k+1)\\pi/2 : k\\in\\mathbb{Z}\\}$, matching option B.

📖 Explanation: Because cosine is continuous, the image of any sequence converging to $c$ must converge to \\\cos c\. Therefore the sequence \\\cos x_n\ converges to \\\cos c\. Option A expresses this conclusion.

📖 Explanation: At \x = \\pi\, \\\tan \\pi = 0\ because \\\sin \\pi = 0\ and \\\cos \\pi = -1\\neq 0\; thus \\\tan x\ is continuous there. Conversely, \\\cot x = \\cos x / \\sin x\ is undefined because \\\sin \\pi = 0\. Hence only \\\tan x\ remains continuous, which is option C.

📖 Explanation: The statement \\\lim_{x\\to \\pi/4}\\sin x = \\sin(\\pi/4)\ directly reflects the definition of continuity for \\\sin x\ at the point \c = \\pi/4\. This is the correct illustration, corresponding to option A.

📖 Explanation: Rewrite \1-\\cos(2x)=2\\sin^2 x\. Then \\\frac{1-\\cos(2x)}{x^2}=2\\left(\\frac{\\sin x}{x}\\right)^2\. As \x\\to 0\, \\\frac{\\sin x}{x}\\to 1\, so the limit equals \2\\cdot1^2 = 2\. Option C provides this value.

📖 Explanation: On the closed interval \[0,\\pi/4]\ the secant function is continuous and its domain contains no points where \\\cos x = 0\. By the Heine–Cantor theorem, any continuous function on a closed, bounded interval is uniformly continuous. Hence option A is correct.