Q.Prove that the function $f(x) = 5x - 3$ is continuous at $x = 0$, at $x = -3$ and at $x = 5$.

Showing the 12 most recent of 32 on this concept.

1. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.The function $f(x)=\begin{cases}\dfrac{2}{5-x}, & x<3\\ 5-x, & x \ge 3\end{cases}$ is (A) left discontinuous at $x=3$ (B) left continuous at $x=3$ (C) right discontinuous at $x=5$ (D) discontinuous at $x=5$

   ›Reveal solutionSolution

   Checking the one-sided limits at the junction $x=3$ shows the function jumps from $1$ to $2$ on approach from the left, i.e. it is left-discontinuous (but right-continuous) at $x=3$; there is nothing special happening at $x=5$.

   Concept and Intuition

   A piecewise function can fail to be continuous from only one side at a junction point if the junction point itself is included in one branch. Here $x=3$ is the boundary, and it belongs to the $5-x$ branch (since that branch is defined for $x\ge3$), so we must check whether the other branch's limit (from the left) agrees with the actual function value.

   Step-by-Step Solution

   1. Function value at $x=3$: since $x\ge3$ uses $f(x)=5-x$, $f(3)=5-3=2$.
   2. Left-hand limit as $x\to3^-$: uses $f(x)=\dfrac{2}{5-x}$, so $\lim_{x\to3^-}\dfrac{2}{5-x}=\dfrac{2}{5-3}=1$.
   3. Since LHL $=1\ne f(3)=2$, the function is discontinuous when approached from the left — i.e. left discontinuous at $x=3$.
   4. Right-hand limit as $x\to3^+$: still $5-x\to2=f(3)$, so it is right-continuous at $x=3$ — this rules out "left continuous" (option B, false).
   5. At $x=5$: for all $x\ge3$ (which includes a neighbourhood of $5$), $f(x)=5-x$ is just a polynomial — continuous everywhere. So there is no discontinuity of any kind at $x=5$, ruling out options (C) and (D).

   Common Mistakes

   • Assuming the "other" piece ($2/(5-x)$) is relevant near $x=5$ — it only applies for $x<3$, nowhere near $5$.
   • Confusing "left discontinuous" with "discontinuous from the right" — here the right side actually matches the true value.

   ✓Final answer

   The correct option is (A) — left discontinuous at $x=3$.

   ANSWER: A

2. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.If the function $f(x) = \dfrac{\sqrt{1+x}-1}{x}$ is continuous at $x = 0$ then $f(0) =$ (A) $-\dfrac{1}{2}$ (B) $\dfrac{1}{3}$ (C) $\dfrac{1}{2}$ (D) $-\dfrac{1}{3}$

   ›Reveal solutionSolution

   Rationalizing the numerator turns the $0/0$ form into a simple limit, which must equal $f(0)$ for continuity.

   Concept and Intuition

   For $f$ to be continuous at $x=0$, we need $f(0)=\displaystyle\lim_{x\to0}f(x)$. Since the given formula for $f(x)$ is $0/0$ at $x=0$, we must simplify it algebraically first.

   Step-by-Step Solution

   1. $\dfrac{\sqrt{1+x}-1}{x} = \dfrac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x(\sqrt{1+x}+1)} = \dfrac{(1+x)-1}{x(\sqrt{1+x}+1)} = \dfrac{x}{x(\sqrt{1+x}+1)} = \dfrac{1}{\sqrt{1+x}+1}$.
   2. As $x\to0$: $\dfrac{1}{\sqrt{1}+1} = \dfrac12$.
   3. For continuity at $x=0$: $f(0) = \dfrac12$.

   Common Mistakes

   • Forgetting to rationalize and instead trying to plug $x=0$ directly (giving an indeterminate $0/0$).
   • Sign error in the conjugate multiplication.

   ✓Final answer

   The correct option is (C) — $\dfrac{1}{2}$.

   ANSWER: C

3. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.If a real valued function $f(x)=\begin{cases}\dfrac{x^2+(a+3)x+(a+1)}{x+3}, & x\neq -3\\ -\dfrac{5}{2}, & x=-3\end{cases}$ is continuous at $x=-3$, then $\displaystyle\lim_{x\to a}(x^2+x+1) =$ (A) $\dfrac{7}{4}$ (B) $\dfrac{5}{2}$ (C) $\dfrac{4}{7}$ (D) $\dfrac{2}{5}$

   ›Reveal solutionSolution

   This tests continuity of a rational function with a removable-type discontinuity: matching the limit to the given value pins down the parameter $a$, then the asked limit is a trivial polynomial evaluation. Answer: $\dfrac{7}{4}$.

   Concept and Intuition

   A piecewise function is continuous at a point if the limit of the "elsewhere" formula, as $x$ approaches that point, equals the value assigned at that point. Here the formula is a rational function whose denominator vanishes at $x=-3$; for the limit to exist (and be finite, matching $-\tfrac52$), the numerator must also vanish there, so that $(x+3)$ cancels — this is the standard "$\frac00$ removable singularity" idea.

   Step-by-Step Solution

   1. Continuity at $x=-3$ requires $\displaystyle\lim_{x\to -3}\frac{x^2+(a+3)x+(a+1)}{x+3} = -\frac52$.
   2. Since the denominator $\to 0$, the numerator must vanish at $x=-3$ (else the limit is $\pm\infty$, not finite):

   $(-3)^2+(a+3)(-3)+(a+1) = 9-3a-9+a+1 = 1-2a = 0 \implies a=\frac12.$

   3. Check: with $a=\tfrac12$, numerator $=x^2+\tfrac72x+\tfrac32$. Factor out $(x+3)$: $x^2+\tfrac72x+\tfrac32 = (x+3)\left(x+\tfrac12\right)$ (verify: $(x+3)(x+\tfrac12)=x^2+\tfrac12x+3x+\tfrac32=x^2+\tfrac72x+\tfrac32$ ✓).
   4. So $\displaystyle\lim_{x\to -3}\frac{(x+3)(x+\frac12)}{x+3} = -3+\frac12 = -\frac52$, matching the given value — confirming $a=\tfrac12$.
   5. Now evaluate $\displaystyle\lim_{x\to a}(x^2+x+1) = \lim_{x\to 1/2}(x^2+x+1)$. Since $x^2+x+1$ is a polynomial (continuous everywhere), the limit equals direct substitution:

   $\left(\frac12\right)^2+\frac12+1 = \frac14+\frac12+1 = \frac{1+2+4}{4}=\frac74.$

   Common Mistakes

   • Forgetting that a finite limit at a point where the denominator vanishes forces the numerator to vanish too (skipping this step leads to an unsolvable/incorrect $a$).
   • Confusing "$\lim_{x\to a}$" with "$\lim_{x\to -3}$" — here $a=\tfrac12$ is just a number, so the second limit is a plain evaluation, not another continuity condition.

   ✓Final answer

   The correct option is (A) — $\dfrac{7}{4}$.

   ANSWER: A

4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If $f(x) = \begin{cases} \left(\frac{2+3x^2}{2+5x^2}\right)^{\frac{8}{x^2}+3}, & \text{for } x \neq 0 \\ k, & \text{for } x = 0 \end{cases}$ is a continuous function at $x = 0$, then $k =$ (A) $e^{-2}$ (B) $e^{-4}$ (C) $e^{-8}$ (D) $e^{-16}$

   ›Reveal solutionSolution

   This tests the standard $1^\infty$ exponential limit technique. Taking logs and expanding to first order in $x^2$ gives $k=e^{-8}$.

   Concept and Intuition

   Expressions of the form (base $\to1$)$^{\text{power}\to\infty}$ are evaluated by writing $L=e^{\lim(\text{power})\cdot\log(\text{base})}$ and expanding $\log(1+u)\approx u$ for the small quantity $u=$ base$-1$. The constant added to the exponent (here $+3$) never survives, since it multiplies a term that itself $\to0$.

   Step-by-Step Solution

   1. For continuity at $x=0$: $k=\lim_{x\to0}\left(\frac{2+3x^2}{2+5x^2}\right)^{8/x^2+3}$.
   2. Let $L$ denote this limit. Then $\log L=\lim_{x\to0}\left(\frac{8}{x^2}+3\right)\log\left(\frac{2+3x^2}{2+5x^2}\right)$.
   3. Write the base as $1+u$ where $u=\frac{2+3x^2}{2+5x^2}-1=\frac{-2x^2}{2+5x^2}\approx -x^2$ for small $x$ (to leading order).
   4. $\log(1+u)\approx u\approx -x^2$ for small $x$.
   5. So $\left(\frac{8}{x^2}+3\right)(-x^2)=-8-3x^2\to -8$ as $x\to0$ (the $-3x^2$ term vanishes, and the correction beyond leading order in $u$ also vanishes since it's multiplied by $x^2$ then divided again giving $O(x^2)$).
   6. So $\log L=-8\Rightarrow L=e^{-8}=k$.

   Common Mistakes

   • Forgetting the $+3$ term but assuming it changes the answer — it drops out because it is multiplied by a quantity that itself tends to zero.
   • Using $u=\frac{-2x^2}{2}=-x^2$ without checking that higher order corrections vanish in the final limit.

   ✓Final answer

   The correct option is (C) — $e^{-8}$.

   ANSWER: C

5. 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.
   3. At $x=2$: $f(2)=-\dfrac32$ (from the $x\ge2$ branch). Left-hand limit: $\lim_{x\to2^-}\left(\dfrac12-x\right)=\dfrac12-2=-\dfrac32$. Right-hand limit (same branch, $x\ge2$): $-\dfrac32$.
   4. All three values equal $-\dfrac32$, so $f$ is continuous at $x=2$ — (D) is true.

   Common Mistakes

   • Confusing which piece of the definition applies right at the boundary point itself (e.g. using the open-interval piece instead of the explicitly given value at $x=1$ or $x=2$).
   • Mixing up left/right limits when checking one-sided continuity.

   ✓Final answer

   The correct option is (D) — $f$ is continuous at $x = 2$.

   ANSWER: D

6. 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. $e^{a}=e^{2/3}=b$.
   4. Matching exponents: $a=\dfrac23$, and correspondingly $b=e^{2/3}$.

   Common Mistakes

   • Not recognizing the $(1+u)^{1/u}\to e$ form and instead trying to directly evaluate $u^{a/u}$ (which behaves completely differently — it vanishes, losing the $a$-dependence entirely).
   • Small-angle slip: using $\tan kx\approx kx$ only for the numerator or only for the denominator.

   ✓Final answer

   The correct option is (B) — $a=\dfrac23,\ b=e^{2/3}$.

   ANSWER: B

7. 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 $x\to0$.
   5. For $f$ to be continuous at $x=0$, we must define $f(0)=\displaystyle\lim_{x\to0}f(x)=1$.

   Common Mistakes

   • Using $\tan x\sim x$ but then forgetting to also apply $\log_e(1+u)\sim u$, leading to an incorrect order-of-magnitude comparison.
   • Mixing up $\sin(x^3)$ with $(\sin x)^3$ — here it is $\sin$ evaluated at $x^3$, which is still $\sim x^3$ for small $x$.

   ✓Final answer

   The correct option is (A) — $1$.

   ANSWER: A

8. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.$f(x)=\begin{cases}\dfrac{(2x^2-ax+1)-(ax^2+3bx+2)}{x+1}, & \text{if } x\neq -1\\ k, & \text{if } x=-1\end{cases}$ is a real valued function. If $a,b,k\in R$ and $f$ is continuous on $R$ then $k=$ (A) $-\dfrac{1}{3}$ (B) 6 (C) $a-2$ (D) $a-3$

   ›Reveal solutionSolution

   Continuity at the removable-type point requires the numerator to vanish there (fixing $b$) and then simplifying the quotient to evaluate $k$ as a limit. The answer is $k=a-3$.

   Concept and Intuition

   A rational expression $\dfrac{N(x)}{x+1}$ can only have a finite limit as $x\to-1$ if $N(-1)=0$ (otherwise the limit is $\pm\infty$ and no choice of $k$ can make $f$ continuous there). So the first job is to use that vanishing condition to pin down any free constant, then simplify by cancelling the common factor.

   Step-by-Step Solution

   1. Numerator: $(2x^2-ax+1)-(ax^2+3bx+2)=(2-a)x^2-(a+3b)x-1$.
   2. At $x=-1$: $(2-a)(1)+(a+3b)-1=1+3b$. For the limit (hence continuity) to exist finitely, this must be $0$: $b=-\dfrac13$.
   3. With $b=-1/3$, numerator becomes $(2-a)x^2-(a-1)x-1$.
   4. Factor out $(x+1)$: writing $(2-a)x^2-(a-1)x-1=(x+1)\left[(2-a)x-1\right]$ (verified by expansion, matching all coefficients).
   5. So for $x\ne-1$: $f(x)=\dfrac{(x+1)[(2-a)x-1]}{x+1}=(2-a)x-1$.
   6. Continuity requires $k=\lim_{x\to-1}f(x)=(2-a)(-1)-1=-(2-a)-1=a-3$.

   Common Mistakes

   • Forgetting that the numerator must vanish at $x=-1$ before the limit can even exist — jumping straight to L'Hôpital without checking this.
   • Sign errors while combining $-(2-a)-1$.

   ✓Final answer

   The correct option is (D) — $a-3$.

   ANSWER: D

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

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

   3. Match for continuity: Since $f$ is continuous at $0$, $f(0)$ must equal both one-sided limits:

   $f(0) = \frac12$

   Common Mistakes

   • Forgetting to rationalize/simplify the $\sqrt{x^2+x}-\sqrt{x}$ term before taking the limit, leading to an indeterminate form that seems to diverge.
   • Using only the first-order term of $\sin t \approx t$ without checking the higher-order terms vanish appropriately (they do, since we only need the leading behavior).

   ✓Final answer

   The correct option is (A) — $1/2$.

   ANSWER: A

10. 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.
    6. Among the given points $0,\,1/3,\,1,\,-1$, only $x=1/3$ is where $f$ fails to be continuous (finite).

    Common Mistakes

    • Thinking $f(x)=1/(1-3x)$ is continuous everywhere in its natural formula domain, forgetting the series itself only converges (and defines $f$) up to $x=1/3$.
    • Confusing this with $x=1$, which would be the boundary if the ratio were simply $x$ instead of $3x$.

    ✓Final answer

    The correct option is (B) — $\frac{1}{3}$.

    ANSWER: B

11. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
    Q.If a function $f(x)=\begin{cases}\dfrac{\tan((\alpha+1)x)+\tan2x}{x} & \text{if } x>0\\ \beta & \text{at } x=0\\ \dfrac{\sin3x-\tan3x}{x^3} & \text{if } x<0\end{cases}$ is continuous at $x=0$ then $|\alpha|+|\beta|=$ (A) $60$ (B) $30$ (C) $15$ (D) $45$

    ›Reveal solutionSolution

    Continuity at $x=0$ forces both one-sided limits to equal $\beta=f(0)$; compute each limit via small-angle expansions and solve for $\alpha,\beta$.

    Concept and Intuition

    For a piecewise function to be continuous at a point, the left-hand limit, the right-hand limit, and the function's value there must all agree. Here both one-sided limits are $0/0$-type indeterminate forms requiring standard small-$x$ expansions of $\tan$ and $\sin$.

    Step-by-Step Solution

    1. Right-hand limit ($x\to0^+$): using $\lim_{x\to0}\tan(kx)/x=k$, $\lim_{x\to0^+}\frac{\tan((\alpha+1)x)+\tan2x}{x}=(\alpha+1)+2=\alpha+3.$ This must equal $\beta$: $\beta=\alpha+3$.
    2. Left-hand limit ($x\to0^-$): expand $\sin3x\approx3x-\dfrac{(3x)^3}{6}=3x-4.5x^3$ and $\tan3x\approx3x+\dfrac{(3x)^3}{3}=3x+9x^3$.
    3. $\sin3x-\tan3x\approx(3x-4.5x^3)-(3x+9x^3)=-13.5x^3=-\dfrac{27}{2}x^3$.
    4. So $\lim_{x\to0^-}\dfrac{\sin3x-\tan3x}{x^3}=-\dfrac{27}{2}$. This must also equal $\beta$: $\beta=-\dfrac{27}{2}$.
    5. From step 1: $\alpha=\beta-3=-\dfrac{27}{2}-3=-\dfrac{33}{2}$.
    6. $|\alpha|+|\beta|=\dfrac{33}{2}+\dfrac{27}{2}=\dfrac{60}{2}=30$.

    Common Mistakes

    • Using $\tan u\approx u+u^3/3$ but forgetting to cube the coefficient properly (using $(3x)^3=27x^3$, not $3x^3$).
    • Sign confusion between the $\sin$ and $\tan$ expansions' cubic-term coefficients ($-1/6$ vs $+1/3$).

    ✓Final answer

    The correct option is (B) — $30$.

    ANSWER: B

12. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
    Q.If a 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) $8$ (B) $4$ (C) $3$ (D) $2$

    ›Reveal solutionSolution

    Both one-sided limits at $x=0$ evaluate to $8$ (via $1-\cos\theta=2\sin^2(\theta/2)$ on the left, and rationalizing on the right), so $a=8$.

    Concept and Intuition

    For $f$ to be continuous at $0$, the left-hand limit, right-hand limit, and $f(0)=a$ must all agree. The left piece is a classic $0/0$ trig limit solved via the half-angle identity for $1-\cos\theta$; the right piece is a classic surd limit solved by rationalizing the denominator.

    Step-by-Step Solution

    1. Left limit: $1-\cos4x=2\sin^2(2x)$, so $\dfrac{1-\cos4x}{x^2}=\dfrac{2\sin^2(2x)}{x^2}=8\left(\dfrac{\sin2x}{2x}\right)^2$.
    2. As $x\to0^-$, $\dfrac{\sin2x}{2x}\to1$, so this limit $\to8(1)^2=8$.
    3. Right limit: let $t=\sqrt x$ (so $t\to0^+$ as $x\to0^+$): $\dfrac{\sqrt x}{\sqrt{16+\sqrt x}-4}=\dfrac{t}{\sqrt{16+t}-4}$.
    4. Rationalize: multiply by $\dfrac{\sqrt{16+t}+4}{\sqrt{16+t}+4}$: $\dfrac{t(\sqrt{16+t}+4)}{(16+t)-16}=\dfrac{t(\sqrt{16+t}+4)}{t}=\sqrt{16+t}+4$.
    5. As $t\to0^+$: $\sqrt{16+0}+4=4+4=8$.
    6. Both one-sided limits equal $8$; continuity at $x=0$ requires $f(0)=a=8$.

    Common Mistakes

    • Forgetting the factor of $2$ inside $\left(\dfrac{\sin2x}{2x}\right)^2$ and getting $2$ instead of $8$ for the left limit.
    • Not substituting $t=\sqrt x$ and instead trying to rationalize directly in $x$, which is messier and error-prone.

    ✓Final answer

    The correct option is (A) — $8$.

    ANSWER: A