Q.Examine the continuity of $f$, where $f$ is defined by $f(x) = \begin{cases} \sin x - \cos x, & \text{if } x \neq 0 \\ -1, & \text{if } x = 0 \end{cases}$

Showing the 12 most recent of 32 on this concept.

1. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.If $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=\begin{cases} \dfrac{\sin x - \sin\dfrac{x}{2}}{x}, & x<0 \\[2mm] \dfrac{\sqrt{x^2+x}-\sqrt{x}}{x^{3/2}}, & x>0 \end{cases}$ is continuous on $\mathbb{R}$, then $f(0)=$ (A) $1/2$ (B) $3/2$ (C) $1$ (D) $-1$

   ›Reveal solutionSolution

   This tests continuity of a piecewise function at the junction point using standard small-$x$ approximations; both one-sided limits equal $1/2$.

   Concept and Intuition

   A piecewise function is continuous at a boundary point only if the left-hand limit, right-hand limit, and the defined value at that point all agree. Here $f(0)$ isn't given directly by either branch (both blow up as $0/0$), so we must compute the limits from each side and set $f(0)$ equal to their common value.

   Step-by-Step Solution

   1. Left-hand limit ($x\to 0^-$):

   $\lim_{x\to0^-}\frac{\sin x-\sin(x/2)}{x}$

   Use the small-angle expansion $\sin t = t - \dfrac{t^3}{6}+\cdots$:

   $\sin x - \sin\frac{x}{2} = \left(x-\frac{x^3}{6}\right)-\left(\frac{x}{2}-\frac{x^3}{48}\right)+\cdots = \frac{x}{2} + O(x^3)$

   Dividing by $x$: the limit is $\dfrac12$.

   2. Right-hand limit ($x\to 0^+$):

   $\lim_{x\to0^+}\frac{\sqrt{x^2+x}-\sqrt{x}}{x^{3/2}} = \lim_{x\to0^+}\frac{\sqrt{x}\left(\sqrt{x+1}-1\right)}{x^{3/2}} = \lim_{x\to0^+}\frac{\sqrt{x+1}-1}{x}$

   Using $\sqrt{1+x}\approx 1+\dfrac{x}{2}-\dfrac{x^2}{8}+\cdots$:

   $\frac{\sqrt{x+1}-1}{x}\approx \frac{x/2}{x} = \frac12$ …

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2. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If $f(x) = \dfrac{\cos^4 x - 1}{x^2}$, $x \ne 0$ and $f(0) = 2$ is a real valued function, then (A) $\displaystyle\lim_{x \to 0} f(x)$ does not exist (B) $\displaystyle\lim_{x \to 0} f(x) = 1$ (C) f is not continuous at x = 0 (D) f is continuous at x = 0

   ›Reveal solutionSolution

   This tests factoring $\cos^4x-1$ and using the standard limit $\frac{\sin x}{x}\to1$; the limit exists and equals $-2$, but $f(0)=2$, so $f$ fails continuity at $0$.

   Concept and Intuition

   $\cos^4x-1$ factors as a difference of squares twice: $\cos^4x-1=(\cos^2x-1)(\cos^2x+1)=-\sin^2x(\cos^2x+1)$. This lets us isolate the familiar $\left(\frac{\sin x}{x}\right)^2\to1$ building block.

   Step-by-Step Solution

   1. $\cos^4x - 1 = (\cos^2x-1)(\cos^2x+1) = -\sin^2x\,(\cos^2x+1)$.
   2. $f(x) = \dfrac{-\sin^2x(\cos^2x+1)}{x^2} = -\left(\dfrac{\sin x}{x}\right)^2(\cos^2x+1)$.
   3. As $x\to0$: $\left(\dfrac{\sin x}{x}\right)^2\to1$ and $\cos^2x+1\to2$.
   4. So $\displaystyle\lim_{x\to0}f(x) = -1\times2 = -2$ — this limit exists (rules out (A)) and is not $1$ (rules out (B)). …

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3. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If a real valued function $f(x) = \begin{cases} (1+\sin x)^{\operatorname{cosec}x}, & -\pi/2 < x < 0 \\ a, & x = 0 \\ \dfrac{e^{2/x}+e^{3/x}}{ae^{2/x}+be^{3/x}}, & 0 < x < \pi/2 \end{cases}$ is continuous at x = 0, then $ab =$ (A) $e$ (B) $e^2$ (C) 1 (D) $-1$

   ›Reveal solutionSolution

   The left-hand limit is the classical $1^\infty$ form giving $e$, fixing $a=e$;

   the right-hand limit needs dividing by the dominant exponential to fix $b$.

   Continuity forces both, and $ab=1$.

   Concept and Intuition

   A piecewise function is continuous at a point only if the left-hand limit,

   right-hand limit, and the function's value there all agree. Here the left

   piece is a $1^\infty$ indeterminate form (standard trick: exponentiate and use

   $\log(1+u)\sim u$), and the right piece is a ratio of two exponentials growing at

   different rates as $x\to0^+$ (since $1/x\to+\infty$), so the faster-growing

   exponential $e^{3/x}$ dominates and everything else becomes negligible after

   dividing through by it.

   Step-by-Step Solution

   1. Left limit: $L=\lim_{x\to0^-}(1+\sin x)^{\operatorname{cosec}x}$. Take logs: $\log L=\lim \operatorname{cosec}x\cdot\log(1+\sin x)=\lim\dfrac{\sin x}{\sin x}\left(1+O(\sin x)\right)\to1$. So $L=e$. Continuity requires $f(0)=a=L=e$.
   2. Right limit: $R=\lim_{x\to0^+}\dfrac{e^{2/x}+e^{3/x}}{ae^{2/x}+be^{3/x}}$. Divide numerator and denominator by $e^{3/x}$: $R=\lim_{x\to0^+}\dfrac{e^{-1/x}+1}{ae^{-1/x}+b}$. …

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4. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.Which one of the following function is discontinuous at x=1? (A) $f(x)=\sin^2x+\tan^2x+\cos^2x-\sec^2x$ (B) $f(x)=\dfrac{1}{1+2^{\sin x}}$ (C) $f(x)=\begin{cases}\dfrac{x-1}{|x-1|+2(x-1)^2}, & x\neq1\\ 1, & x=1\end{cases}$ (D) $f(x)=e^{x}+5$

   ›Reveal solutionSolution

   Options (A), (B), (D) simplify to functions that are continuous everywhere; option (C)'s one-sided limits at $x=1$ disagree ($+1$ from the right, $-1$ from the left), so it alone is discontinuous at $x=1$.

   Concept and Intuition

   For piecewise/rational-looking expressions, check whether the identity $\sin^2x+\cos^2x=1$ and $\sec^2x-\tan^2x=1$ collapse them to constants (continuous), whether a denominator can vanish, and — for absolute-value expressions — whether the left and right limits genuinely agree.

   Step-by-Step Solution

   1. (A) $f(x)=\sin^2x+\tan^2x+\cos^2x-\sec^2x=(\sin^2x+\cos^2x)-(\sec^2x-\tan^2x)=1-1=0$ for all $x$ where $\tan,\sec$ are defined; at $x=1$ (radian) $\cos1\ne0$, so it's fine and continuous ($\equiv0$ near $x=1$).
   2. (B) $f(x)=\dfrac{1}{1+2^{\sin x}}$: since $2^{\sin x}>0$ always, the denominator never vanishes ($\ge1+2^{-1}=1.5$); continuous everywhere, including $x=1$.
   3. (C) For $x\ne1$, let $t=x-1$: $f=\dfrac{t}{|t|+2t^2}$.
      • As $t\to0^+$ ($x\to1^+$): $|t|=t$, so $f=\dfrac{t}{t+2t^2}=\dfrac{1}{1+2t}\to1$.
      • As $t\to0^-$ ($x\to1^-$): $|t|=-t$, so $f=\dfrac{t}{-t+2t^2}=\dfrac{1}{-1+2t}\to-1$. …

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5. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.If $f(x) = \dfrac{\log_e\left(1 + x^2(\tan x)\right)}{\sin x^3}$, $x \neq 0$ is to be continuous at $x = 0$, then $f(0)$ must be equal to ______ (A) $1$ (B) $0$ (C) $\dfrac{1}{2}$ (D) $-1$

   ›Reveal solutionSolution

   Using the small-angle equivalences $\log_e(1+u)\sim u$, $\tan x\sim x$, and $\sin x^3\sim x^3$ near $x=0$, the limit of $f(x)$ works out to $1$, so $f(0)=1$ is required for continuity.

   Concept and Intuition

   For a function to be continuous at a removable-looking point like $x=0$, its value there must be defined equal to the limit of the function as $x\to0$. Standard small-$x$ equivalences ($\log_e(1+u)\sim u$ for small $u$; $\tan x\sim x$; $\sin x\sim x$) let us evaluate the $\frac{0}{0}$-type limit cleanly.

   Step-by-Step Solution

   1. As $x\to0$, let $u=x^2\tan x$. Since $\tan x\sim x$ for small $x$, $u\sim x^2\cdot x=x^3\to0$.
   2. Using $\log_e(1+u)\sim u$ for small $u$: $\log_e(1+x^2\tan x)\sim x^2\tan x\sim x^3$.
   3. For the denominator, $\sin(x^3)\sim x^3$ as $x\to0$ (since $\sin\theta\sim\theta$ for small $\theta$, here $\theta=x^3\to0$).
   4. So $f(x)=\dfrac{\log_e(1+x^2\tan x)}{\sin(x^3)}\to\dfrac{x^3}{x^3}=1$ as $x\to0$. …

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6. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.Let $f(x) = \begin{cases} 0, & x = 0 \\ 2 - x, & \text{for } 0 < x < 1 \\ 2, & \text{for } x = 1 \\ \dfrac{1}{2} - x, & \text{for } 1 < x < 2 \\ \dfrac{-3}{2}, & \text{for } x \ge 2 \end{cases}$ then which of the following is true (A) $f$ is right continuous at $x = 0$ (B) $f$ is left continuous at $x = 1$ (C) $f$ is right continuous at $x = 1$ (D) $f$ is continuous at $x = 2$

   ›Reveal solutionSolution

   Compute the left-hand limit, right-hand limit, and function value at each candidate point and compare; only at $x=2$ do all three coincide.

   Concept and Intuition

   A function is continuous at a point if the left-hand limit, right-hand limit, and the function's value there all agree. "Right continuous" only needs the right-hand limit to equal the function value; "left continuous" only needs the left-hand limit to match.

   Step-by-Step Solution

   1. At $x=0$: $f(0)=0$. Right-hand limit: $\lim_{x\to0^+}(2-x)=2$. Since $2\ne0$, $f$ is not right continuous at $0$ — (A) is false.
   2. At $x=1$: $f(1)=2$. Left-hand limit: $\lim_{x\to1^-}(2-x)=1$. Since $1\ne2$, not left continuous — (B) is false. Right-hand limit: $\lim_{x\to1^+}\left(\dfrac12-x\right)=-\dfrac12$. Since $-\dfrac12\ne2$, not right continuous either — (C) is false. …

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7. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If $f(x)$ defined by $f(x) = \begin{cases} \dfrac{1 - \tan x}{4x - \pi}, & x \neq \dfrac{\pi}{4}, x \in \left[0, \dfrac{\pi}{2}\right] \\ k, & x = \dfrac{\pi}{4} \end{cases}$ is continuous in $\left[0, \dfrac{\pi}{2}\right]$, then $k =$ (A) $-1$ (B) $-\dfrac{1}{2}$ (C) $\dfrac{1}{2}$ (D) $1$

   ›Reveal solutionSolution

   Substituting $x=\pi/4+h$ and using the tangent subtraction identity, the limit of

   $\dfrac{1-\tan x}{4x-\pi}$ as $x\to\pi/4$ works out to $-\dfrac12$, which is the required

   value of $k$.

   Concept and Intuition

   For continuity at $x=\pi/4$, we need $k=\lim_{x\to\pi/4}f(x)$. Since the expression is $0/0$

   at $x=\pi/4$, shifting variables via $x=\pi/4+h$ (so $h\to0$) turns $\tan x$ into

   $\tan(\pi/4+h)$, which has a clean expansion using the tangent-addition formula.

   Step-by-Step Solution

   1. Let $x=\dfrac{\pi}{4}+h$, so $h\to0$ as $x\to\pi/4$.
   2. $\tan\left(\dfrac{\pi}{4}+h\right)=\dfrac{1+\tan h}{1-\tan h}$.
   3. $1-\tan x = 1-\frac{1+\tan h}{1-\tan h}=\frac{(1-\tan h)-(1+\tan h)}{1-\tan h}=\frac{-2\tan h}{1-\tan h}$
   4. $4x-\pi = 4\left(\dfrac{\pi}{4}+h\right)-\pi = 4h$.
   5. So

   $f(x)=\frac{-2\tan h}{4h(1-\tan h)}$

   6. As $h\to0$, $\tan h\sim h$, so …

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8. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.Let $S_n=1+3x+9x^2+27x^3+\ldots n$ terms and $-\frac{1}{3}<x<\frac{1}{3}$. If $\lim_{n\to\infty}S_n=f(x)$, then $f(x)$ is discontinuous at the point $x=$ (A) $0$ (B) $\frac{1}{3}$ (C) $1$ (D) $-1$

   ›Reveal solutionSolution

   The infinite geometric series sums to $f(x)=1/(1-3x)$ on $(-1/3,1/3)$, and this function has an infinite discontinuity exactly at $x=1/3$, the edge of the interval where the series stops converging.

   Concept and Intuition

   An infinite geometric series $1+r+r^2+\dots$ converges to $\frac{1}{1-r}$ only when $|r|<1$; as $r\to1$ this sum diverges to infinity. Here $r=3x$, so the series converges for $|x|<1/3$ and the resulting closed-form function has a vertical asymptote right at the boundary $x=1/3$.

   Step-by-Step Solution

   1. $S_n = 1+3x+(3x)^2+\dots+(3x)^{n-1}$, a geometric series with first term $1$ and common ratio $3x$.
   2. Sum: $S_n=\dfrac{1-(3x)^n}{1-3x}$ (for $3x\ne1$).
   3. For $-\tfrac13<x<\tfrac13$, $|3x|<1$, so $(3x)^n\to0$ as $n\to\infty$.
   4. Hence $f(x)=\lim_{n\to\infty}S_n = \dfrac{1}{1-3x}$.
   5. This function $f(x)=\frac{1}{1-3x}$ is undefined/blows up exactly where $1-3x=0$, i.e. $x=\tfrac13$ — an infinite discontinuity right at the edge of the domain of convergence. …

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9. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.If $f(x) = \begin{cases} \dfrac{x-[x]}{x-2}, & x>2 \\ b, & x=2 \\ \dfrac{|x^2-x-2|}{a(2+x-x^2)}, & -1<x\le 2 \\ 2a-b, & x\le -1 \end{cases}$ is continuous on $\mathbb{R}$, then $\displaystyle\lim_{x\to 0}\dfrac{\sin^2 ax + x\tan bx}{x^2} =$ (A) 0 (B) 1 (C) 2 (D) 3

   ›Reveal solutionSolution

   Continuity of the piecewise function pins down $a=1,b=1$; substituting these into the limit expression and using standard small-angle limits gives $2$.

   Concept and Intuition

   For a piecewise function to be continuous at a junction point, the one-sided limits and the defined value there must all agree. Here the junctions at $x=2$ and $x=-1$ give the equations needed to solve for the unknown constants $a,b$ before the actual limit can be evaluated.

   Step-by-Step Solution

   1. Right limit at $x=2$: for $x$ slightly $>2$, $[x]=2$, so $f(x)=\dfrac{x-2}{x-2}=1$. So $\lim_{x\to2^+}f(x)=1$.
   2. Left limit at $x=2$ (third piece): $x^2-x-2=(x-2)(x+1)$ and $2+x-x^2=-(x-2)(x+1)$. For $x$ near $2^-$: $(x-2)<0,(x+1)>0$, so $|x^2-x-2|=(2-x)(x+1)$ and $2+x-x^2=(2-x)(x+1)$ too. So the ratio simplifies to $\dfrac1a$ throughout $(-1,2)$.
   3. Continuity at $x=2$: $1=b=\dfrac1a \Rightarrow a=1,\ b=1$.
   4. Check at $x=-1$: piece 4 value $=2a-b=2-1=1$; piece 3's limit as $x\to-1^+$ is also $\dfrac1a=1$. Consistent ✓. …

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10. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.If the function f defined by $f(x) = \begin{cases} \dfrac{1 - \cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \dfrac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}, & x > 0 \end{cases}$ is continuous at $x = 0$, then $a =$ (A) $1$ (B) $2$ (C) $4$ (D) $8$

    ›Reveal solutionSolution

    Continuity at $x=0$ requires the left limit, the right limit, and $f(0)=a$ to all agree; both one-sided limits work out to $8$. Answer: $a=8$.

    Concept and Intuition

    For a piecewise function to be continuous at a junction point, the value defined there ($a$) must equal both the limit approaching from the left and the limit approaching from the right. Each side here needs a standard trick: the trig side uses the double-angle identity $1-\cos\theta=2\sin^2(\theta/2)$, and the surd side needs rationalisation to remove the $\sqrt{\ }-\sqrt{\ }$ indeterminate form.

    Step-by-Step Solution

    1. Left-hand limit ($x\to0^-$): $1-\cos4x=2\sin^2(2x)$, so

    $\lim_{x\to0^-}\frac{1-\cos4x}{x^2}=\lim_{x\to0^-}\frac{2\sin^2(2x)}{x^2}=2\lim_{x\to0}\left(\frac{\sin2x}{2x}\right)^2\cdot4=2\cdot1\cdot4=8.$

    2. Right-hand limit ($x\to0^+$): rationalise $\dfrac{\sqrt x}{\sqrt{16+\sqrt x}-4}$ by multiplying top and bottom by $\sqrt{16+\sqrt x}+4$: $\frac{\sqrt x\,(\sqrt{16+\sqrt x}+4)}{(16+\sqrt x)-16}=\frac{\sqrt x\,(\sqrt{16+\sqrt x}+4)}{\sqrt x}=\sqrt{16+\sqrt x}+4.$ …

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11. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
    Q.The values of $a$ and $b$ for which the function $f(x)=\begin{cases}1+|\sin x|^{a/|\sin x|}, & \dfrac{-\pi}{6}<x<0 \\ b, & x=0 \\ e^{\tan 2x/\tan 3x}, & 0<x<\dfrac{\pi}{6}\end{cases}$ is continuous at $x=0$ are (A) $a=1, b=\dfrac{2}{3}$ (B) $a=\dfrac{2}{3}, b=e^{2/3}$ (C) $a=\dfrac{2}{3}, b=\dfrac{3}{2}$ (D) $a=-1, b=e^{2/3}$

    ›Reveal solutionSolution

    Both one-sided limits must equal $b$; the right side gives $e^{2/3}$ directly, and the left side (a $(1+u)^{1/u}\to e$-type limit) matches it when $a=2/3$ — (B).

    Concept and Intuition

    Continuity at $x=0$ requires $\displaystyle\lim_{x\to0^-}f(x)=f(0)=\lim_{x\to0^+}f(x)$. The right branch is a standard $e^{\text{ratio of small angles}}$ limit, and the left branch is the classical exponential limit $(1+u)^{1/u}\to e$ as $u\to0$, raised to a power $a$.

    Step-by-Step Solution

    1. Right-hand limit: as $x\to0^+$, $\tan2x\approx2x$ and $\tan3x\approx3x$, so $\dfrac{\tan2x}{\tan3x}\to\dfrac23$. Hence $\displaystyle\lim_{x\to0^+}e^{\tan2x/\tan3x}=e^{2/3}$.
    2. Left-hand limit: let $u=|\sin x|\to0^+$ as $x\to0^-$. The left branch is $(1+u)^{a/u}=\left[(1+u)^{1/u}\right]^{a}$. Since $(1+u)^{1/u}\to e$, this tends to $e^{a}$.
    3. For continuity: left limit $=$ right limit $=f(0)=b$, i.e. $e^{a}=e^{2/3}=b$. …

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12. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
    Q.If $f(x)=\begin{cases}\dfrac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}}, & x\neq -1\\[2mm] \dfrac{1}{\sqrt{\lambda\pi}}, & x=-1\end{cases}$ is right continuous at $x=-1$, then $\lambda=$ (A) $1$ (B) $\pi$ (C) $2\pi$ (D) $2$

    ›Reveal solutionSolution

    Expanding $\cos^{-1}(-1+h)$ near $h=0^+$ and simplifying the resulting $0/0$ form gives the right-hand limit $1/\sqrt{2\pi}$; matching this to $f(-1)=1/\sqrt{\lambda\pi}$ gives $\lambda=2$.

    Concept and Intuition

    Right continuity at $x=-1$ requires $\lim_{x\to-1^+}f(x)=f(-1)$. As $x\to-1^+$, both $\sqrt\pi-\sqrt{\cos^{-1}x}\to0$ and $\sqrt{x+1}\to0$, so we need a careful local expansion of $\cos^{-1}x$ near $x=-1$.

    Step-by-Step Solution

    1. Let $x=-1+h$, $h\to0^+$. Write $\cos^{-1}(-1+h)=\pi-\theta$ where $\theta\to0^+$. Then $\cos(\pi-\theta)=-\cos\theta=-1+h\Rightarrow\cos\theta=1-h$.
    2. For small $\theta$: $\cos\theta\approx1-\theta^2/2$, so $1-\theta^2/2\approx1-h\Rightarrow\theta\approx\sqrt{2h}$.
    3. So $\cos^{-1}x\approx\pi-\sqrt{2h}$, and $\sqrt{\cos^{-1}x}\approx\sqrt\pi\sqrt{1-\sqrt{2h}/\pi}\approx\sqrt\pi\left(1-\dfrac{\sqrt{2h}}{2\pi}\right)=\sqrt\pi-\dfrac{\sqrt{2h}}{2\sqrt\pi}$.
    4. So $\sqrt\pi-\sqrt{\cos^{-1}x}\approx\dfrac{\sqrt{2h}}{2\sqrt\pi}$. …

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