Q.Let $a_1, a_2, \ldots, a_n$ be fixed real numbers and define a function $f(x) = (x - a_1)(x - a_2)\cdots(x - a_n)$. What is $\lim_{x\to a_1} f(x)$? For some $a \neq a_1, a_2, \ldots, a_n$, compute $\lim_{x\to a} f(x)$.

Showing the 12 most recent of 76 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If $\lim\limits_{x \to 0} \dfrac{\tan ax - \sin ax}{3\sin 2x - \sin 6x} = 1$, then $\lim\limits_{x \to a} \dfrac{\log(x-3)}{x-a} =$ (A) $1$ (B) $2$ (C) $3$ (D) $4$

   ›Reveal solutionSolution

   First find $a$ from the given trigonometric limit using small-angle (Taylor) expansions, then evaluate a $\log$ limit at that $a$ using L'Hôpital's rule. Answer: $1$.

   Concept and Intuition

   For $x\to0$ limits of the form $0/0$ built from trig functions, replacing each function by its leading Taylor terms turns the whole limit into a ratio of polynomials in $x$, which is much easier to simplify than repeated L'Hôpital differentiation.

   Step-by-Step Solution

   1. Expand near $u=0$: $\tan u \approx u+\dfrac{u^3}{3}$, $\sin u\approx u-\dfrac{u^3}{6}$. So $\tan u - \sin u \approx \dfrac{u^3}{3}+\dfrac{u^3}{6}=\dfrac{u^3}{2}$. With $u=ax$: numerator $\approx \dfrac{a^3x^3}{2}$.
   2. Expand $\sin2x\approx 2x-\dfrac{(2x)^3}{6}=2x-\dfrac{4x^3}{3}$, so $3\sin2x\approx 6x-4x^3$. Expand $\sin6x\approx 6x-\dfrac{(6x)^3}{6}=6x-36x^3$. Denominator: $3\sin2x-\sin6x \approx (6x-4x^3)-(6x-36x^3)=32x^3$.
   3. So the limit $=\dfrac{a^3x^3/2}{32x^3}=\dfrac{a^3}{64}$. Given this equals $1$: $a^3=64\Rightarrow a=4$. …

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2. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.$\lim\limits_{x \to 0} (\cos 2x)^{\tan x} =$ (A) $0$ (B) $e$ (C) $1$ (D) $e^2$

   ›Reveal solutionSolution

   A $1^\infty$ indeterminate form; taking logs and using small-$x$ approximations shows the exponent-times-log product vanishes, so the limit is $1$.

   Concept and Intuition

   For limits of the form $\lim g(x)^{h(x)}$ where $g(x)\to1$ and $h(x)\to\infty$ (or is unbounded), the standard technique is: let $L$ be the limit, take $\log L = \lim h(x)\log g(x)$, evaluate that simpler limit, then exponentiate back.

   Step-by-Step Solution

   1. Let $L = \displaystyle\lim_{x\to0}(\cos2x)^{\tan x}$. Since $\cos2x\to1$ and $\tan x\to0$, this is $1^0$ actually — but let's verify via logs directly (safe either way).
   2. $\log L = \displaystyle\lim_{x\to0} \tan x \cdot \log(\cos 2x)$.
   3. As $x\to0$: $\tan x \approx x$.
   4. $\cos 2x \approx 1 - \dfrac{(2x)^2}{2} = 1-2x^2$, so using $\log(1+u)\approx u$ for small $u$: $\log(\cos2x)\approx \log(1-2x^2)\approx -2x^2$.
   5. So $\log L \approx \displaystyle\lim_{x\to0} x\cdot(-2x^2) = \lim_{x\to0}(-2x^3) = 0$. …

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3. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If $\displaystyle\lim_{x \to 0} \dfrac{\sin^2 3x \, \tan^4 5x}{x^3 \sin^3 kx} = \dfrac{5}{3}$, then $k =$ (A) $15$ (B) $12$ (C) $25$ (D) $27$

   ›Reveal solutionSolution

   Replacing each trig factor by its small-angle equivalent turns the limit into a pure power ratio, giving $k=15$.

   Concept and Intuition

   For small $x$, $\sin(ax)\sim ax$ and $\tan(ax)\sim ax$ (since $\lim_{u\to0}\frac{\sin u}{u}=\lim_{u\to0}\frac{\tan u}{u}=1$). This lets us replace every trig factor in a limit at $0$ by its linear approximation, turning the whole expression into a simple algebraic power of $x$, provided the overall power of $x$ cancels to give a finite nonzero limit (which it does here, both top and bottom being $x^6$).

   Step-by-Step Solution

   1. $\sin3x\sim3x \Rightarrow \sin^23x\sim9x^2$.
   2. $\tan5x\sim5x \Rightarrow \tan^45x\sim625x^4$.
   3. $\sin kx\sim kx \Rightarrow \sin^3kx\sim k^3x^3$. …

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4. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.$\displaystyle\lim_{x \to 0} \dfrac{e^{3x} - (1 + x)e^{x}}{e^{2x} - 1} =$ (A) $3$ (B) $\dfrac{1}{3}$ (C) $2$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   A quick Taylor expansion (or L'Hôpital) of this $0/0$ form gives the limit $\dfrac12$.

   Concept and Intuition

   Both numerator and denominator vanish at $x=0$, so this is a $0/0$ indeterminate form. Expanding each exponential as a Taylor series around $0$ (or equivalently applying L'Hôpital's rule once) isolates the leading behavior and gives the limit cleanly.

   Step-by-Step Solution

   1. $e^{3x}=1+3x+\dfrac{9x^2}{2}+\dots$
   2. $(1+x)e^x = (1+x)\left(1+x+\dfrac{x^2}{2}+\dots\right) = 1+2x+\dfrac{3x^2}{2}+\dots$
   3. Numerator: $e^{3x}-(1+x)e^x = (1+3x+\dots)-(1+2x+\dots) = x + O(x^2)$.
   4. Denominator: $e^{2x}-1 = 2x+O(x^2)$.
   5. Limit $=\dfrac{x+O(x^2)}{2x+O(x^2)} \to \dfrac{1}{2}$ as $x\to0$.
   6. (Check via L'Hôpital: $f'(x)=3e^{3x}-e^x(2+x)$, $f'(0)=3-2=1$; $g'(x)=2e^{2x}$, $g'(0)=2$; ratio $=1/2$.)

   Common Mistakes …

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5. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If $\lim\limits_{x\to\infty} \dfrac{(ax^2+bx+c)}{(lx^2+mx+n)}\dfrac{(lx-1)^{\frac{x}{2}}}{(lx+a)^{\frac{x}{2}}}=\dfrac{3}{\sqrt{16e}}$, then $\lim\limits_{x\to 0} \dfrac{l+mx+cx^2}{a+bx+nx^2}=$ (A) $\dfrac{3}{4}$ (B) $\dfrac{4}{3}$ (C) $\dfrac{5}{4}$ (D) $\dfrac{4}{5}$

   ›Reveal solutionSolution

   Matching the given exponential-limit value pins down $a=3,\ l=4$; the second limit then evaluates directly (no indeterminate form) to $l/a=4/3$.

   Concept and Intuition

   The given limit splits into a rational-function part (which tends to the ratio of leading coefficients) times an exponential part of the form $\left(1+\frac{k}{x}\right)^{x/2}\to e^{k/2}$. Matching both factors against the target value $\frac{3}{\sqrt{16e}}=\frac{3}{4\sqrt e}$ gives two independent equations for $a$ and $l$. The second requested limit is then a red herring in complexity — at $x=0$ neither the numerator nor denominator vanishes, so it is not an indeterminate form; it's simply $l/a$.

   Step-by-Step Solution

   1. As $x\to\infty$: $\dfrac{ax^2+bx+c}{lx^2+mx+n}\to \dfrac{a}{l}$ (ratio of leading coefficients).
   2. For the exponential factor, write $\dfrac{lx-1}{lx+a} = 1-\dfrac{a+1}{lx+a}$.
   3. As $x\to\infty$, $\left(1-\dfrac{a+1}{lx+a}\right)^{x/2} \to \exp\!\left(-\dfrac{a+1}{l}\cdot\dfrac12\right)=\exp\!\left(-\dfrac{a+1}{2l}\right)$, using $\left(1+\frac{k}{x}\right)^{x}\to e^k$ with $k=-(a+1)/l$ (since $lx+a\sim lx$).
   4. So the full limit is $\dfrac{a}{l}\exp\!\left(-\dfrac{a+1}{2l}\right) = \dfrac{3}{4\sqrt e} = \dfrac34 e^{-1/2}$.
   5. Matching the rational and exponential parts separately: $\dfrac{a}{l}=\dfrac34$ and $\dfrac{a+1}{2l}=\dfrac12 \Rightarrow \dfrac{a+1}{l}=1 \Rightarrow l=a+1$. …

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6. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.$\lim_{x \to \frac{\pi}{2}} \frac{\left(1 - \tan\frac{x}{2}\right)(1 - \sin x)}{\left(1 + \tan\frac{x}{2}\right)(\pi - 2x)^3} =$ (A) $0$ (B) $\frac{1}{8}$ (C) $\frac{1}{16}$ (D) $\frac{1}{32}$

   ›Reveal solutionSolution

   This tests evaluating a $\frac00$ limit near $x=\pi/2$ using the substitution $x=\pi/2+h$ and small-angle expansions. The limit works out to $\frac1{32}$.

   Concept and Intuition

   Near $x=\pi/2$, both the numerator and denominator vanish, so we shift the variable to $h=x-\pi/2\to0$ and expand every trigonometric piece to its leading order in $h$. The tangent half-angle ratio simplifies neatly using the tangent subtraction identity.

   Step-by-Step Solution

   1. Let $x=\frac\pi2+h$, so $h\to0$ as $x\to\frac\pi2$. Then $\pi-2x=\pi-\pi-2h=-2h$, so $(\pi-2x)^3=-8h^3$.
   2. $1-\sin x=1-\sin\left(\frac\pi2+h\right)=1-\cos h\approx\frac{h^2}{2}$ for small $h$.
   3. For $\frac{x}{2}=\frac\pi4+\frac h2$, using $\tan\left(\frac\pi4+\frac h2\right)=\frac{1+\tan(h/2)}{1-\tan(h/2)}$: with $t=\tan(h/2)$, $\frac{1-\tan(x/2)}{1+\tan(x/2)}=\frac{1-\frac{1+t}{1-t}}{1+\frac{1+t}{1-t}}=\frac{-2t}{2}=-t\approx-\frac h2$. …

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7. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.$\lim_{x \to \infty}\left[x\left(\frac{1}{1+x} + \frac{1}{(1+x)(1+2x)} + \frac{1}{(1+2x)(1+3x)} + \cdots + \frac{1}{(1+(n-1)x)(1+nx)}\right)\right] =$ (A) $\frac{1}{1-x}$ (B) $\frac{1}{1+x}$ (C) $1$ (D) $0$

   ›Reveal solutionSolution

   This tests recognizing a telescoping sum inside a limit. After the partial-fraction split, almost every term cancels, and the limit as $x\to\infty$ is simply $1$.

   Concept and Intuition

   Each term $\frac1{(1+kx)(1+(k+1)x)}$ has denominators differing by exactly $x$, which is the classic setup for a partial-fraction telescoping sum: $\frac1{ab}=\frac1{b-a}\left(\frac1a-\frac1b\right)$ when $b-a$ is constant across all terms.

   Step-by-Step Solution

   1. For each $k$, let $a=1+kx$, $b=1+(k+1)x$, so $b-a=x$.
   2. $\frac1{ab}=\frac1{x}\left(\frac1a-\frac1b\right)=\frac1x\left[\frac1{1+kx}-\frac1{1+(k+1)x}\right]$.
   3. Summing from $k=0$ to $n-1$, the sum telescopes: $\sum_{k=0}^{n-1}\frac1{(1+kx)(1+(k+1)x)}=\frac1x\left[\frac1{1+0}-\frac1{1+nx}\right]=\frac1x\left[1-\frac1{1+nx}\right]$.
   4. Multiplying by $x$ (as in the original expression): $x\cdot\frac1x\left[1-\frac1{1+nx}\right]=1-\frac1{1+nx}$. …

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8. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.$\lim_{n \to \infty} \frac{1}{n^2}\sum_{r=1}^{n} r\, e^{r/n} =$ (A) $0$ (B) $1$ (C) $e$ (D) $2e$

   ›Reveal solutionSolution

   The given sum, once written as $\frac1n\sum(r/n)e^{r/n}$, is exactly a Riemann sum for $\int_0^1 xe^x dx$, which evaluates to $1$.

   Concept and Intuition

   When a limit has the shape $\lim_{n\to\infty}\frac1n\sum_{r=1}^n f(r/n)$, it's the definition of the definite integral $\int_0^1 f(x)\,dx$ (Riemann sum with $n$ equal subintervals of $[0,1]$). Recognizing this converts a discrete summation limit into a straightforward integration-by-parts problem.

   Step-by-Step Solution

   1. Rewrite the sum: $\dfrac1{n^2}\sum_{r=1}^n re^{r/n} = \dfrac1n\sum_{r=1}^n\dfrac{r}{n}e^{r/n}$.
   2. This is a Riemann sum for $f(x)=xe^x$ over $[0,1]$, with sample points $x_r=r/n$.
   3. So $\displaystyle\lim_{n\to\infty}\frac1n\sum_{r=1}^n\frac{r}{n}e^{r/n} = \int_0^1 xe^x\,dx$.
   4. Integrate by parts: $\int xe^x dx = xe^x-\int e^x dx = xe^x - e^x+c$. …

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9. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.$\displaystyle\lim_{x \to 2} (x^2 - 3x + 3)^{\frac{1}{x^2 - 4}} =$ (A) $e^{\frac{1}{2}}$ (B) 1 (C) 0 (D) $e^{\frac{1}{4}}$

   ›Reveal solutionSolution

   This tests the standard $1^\infty$ exponential-limit technique; the answer is $e^{1/4}$.

   Concept and Intuition

   When $\lim f(x)=1$ and $\lim g(x)=\pm\infty$ (with the base tending to 1 and the exponent blowing up), $\lim f(x)^{1/g(x)}$ is evaluated using $\lim f(x)^{1/g(x)} = e^{\lim (f(x)-1)/g(x)}$, since $f(x)^{1/g(x)}=e^{\frac{\log f(x)}{g(x)}}$ and $\log f(x)\approx f(x)-1$ near $f=1$.

   Step-by-Step Solution

   1. At $x=2$: base $= 4-6+3=1$, and the exponent $\dfrac{1}{x^2-4}\to\infty$ — a $1^\infty$ form.
   2. $f(x)-1 = x^2-3x+2 = (x-1)(x-2)$.
   3. $g(x) = x^2-4 = (x-2)(x+2)$.
   4. $\dfrac{f(x)-1}{g(x)} = \dfrac{(x-1)(x-2)}{(x-2)(x+2)} = \dfrac{x-1}{x+2}$ for $x\ne2$. …

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10. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
    Q.$\displaystyle\lim_{x \to 0} \dfrac{x \cdot 2^x - x}{1 - \cos x} =$ (A) $\log 2$ (B) $\dfrac{1}{2}\log 2$ (C) $2 \log 2$ (D) $\dfrac{1}{2\log 2}$

    ›Reveal solutionSolution

    Using the standard small-$x$ equivalents $2^x-1\sim x\ln2$ and $1-\cos x\sim x^2/2$, the

    limit evaluates to $2\ln2$.

    Concept and Intuition

    Every $0/0$ limit built from standard functions can be attacked with known small-angle/small-$x$

    equivalents rather than L'Hôpital: $a^x-1\sim x\log a$ and $1-\cos x\sim x^2/2$ as $x\to0$.

    Recognising these turns a seemingly awkward limit into simple algebra.

    Step-by-Step Solution

    1. Rewrite the numerator: $x\cdot2^x-x = x(2^x-1)$.
    2. As $x\to0$: $2^x-1=e^{x\ln2}-1\sim x\ln2$ (first-order Taylor).
    3. So numerator $\sim x\cdot x\ln2 = x^2\ln2$.
    4. Denominator: $1-\cos x\sim \dfrac{x^2}{2}$ as $x\to0$.
    5. Limit: $\lim_{x\to0}\frac{x^2\ln2}{x^2/2}=2\ln2$ …

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11. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
    Q.If $\displaystyle\lim_{x \to \infty} \left(1 + \dfrac{a}{x} + \dfrac{b}{x^2}\right)^{2x} = e^2$, then $a =$ (A) $2$ (B) $1$ (C) $\dfrac{1}{2}$ (D) $0$

    ›Reveal solutionSolution

    Taking logs and using $\log(1+u)\sim u$ for small $u=a/x+b/x^2$, the exponent

    $2x\log(\cdots)\to 2a$ must equal $2$, so $a=1$.

    Concept and Intuition

    Limits of the form $\left(1+\tfrac{\text{(vanishing)}}{\ }\right)^{\text{(growing)}}$ are

    best handled by taking $\log$: if $L=\lim(1+u)^{v}$ with $u\to0,\ v\to\infty$, then

    $\log L=\lim v\cdot u$ (using $\log(1+u)\sim u$), turning an indeterminate exponential form

    into a simple product limit.

    Step-by-Step Solution

    1. Let $L=\displaystyle\lim_{x\to\infty}\left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2x}=e^2$.
    2. Take $\log$: $\log L = \displaystyle\lim_{x\to\infty} 2x\log\left(1+\frac{a}{x}+\frac{b}{x^2}\right)$.
    3. As $x\to\infty$, $u=\dfrac{a}{x}+\dfrac{b}{x^2}\to0$, so $\log(1+u)\sim u-\dfrac{u^2}{2}+\cdots$; only the leading $u\sim a/x$ term survives after multiplying by $2x$ (the $b/x^2$ and $u^2$ terms vanish as $x\to\infty$ once multiplied by $2x$, since they are $O(1/x)$). …

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12. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
    Q.$\displaystyle\lim_{n \to \infty} \dfrac{1}{n}\left[\tan^2\dfrac{\pi}{4n} + \tan^2\dfrac{2\pi}{4n} + \cdots + \tan^2\dfrac{\pi}{4}\right] =$ (A) $1$ (B) $0$ (C) $-1$ (D) $\dfrac{4 - \pi}{\pi}$

    ›Reveal solutionSolution

    Recognize the sum as a Riemann sum over $[0,\pi/4]$ for $\tan^2x$; the limit evaluates to $\dfrac{4-\pi}{\pi}$.

    Concept and Intuition

    A sum of the form $\dfrac1n\sum_{k=1}^n f\!\left(\dfrac{k\pi}{4n}\right)$ is a Riemann sum whose sample points $x_k=\dfrac{k\pi}{4n}$ march from $0$ to $\dfrac\pi4$ in steps of $\Delta x=\dfrac{\pi}{4n}$. Since $\dfrac1n=\dfrac4\pi\cdot\Delta x$, the whole sum converges to $\dfrac4\pi\displaystyle\int_0^{\pi/4}f(x)\,dx$.

    Step-by-Step Solution

    1. Here $f(x)=\tan^2x$, and the sum runs over $x_k=\dfrac{k\pi}{4n}$ for $k=1,\dots,n$ (the last term $\tan^2\frac\pi4$ confirms $x_n=\pi/4$).
    2. Write

    $\frac1n\sum_{k=1}^n\tan^2\!\left(\frac{k\pi}{4n}\right)=\frac4\pi\cdot\frac{\pi}{4n}\sum_{k=1}^n\tan^2\!\left(\frac{k\pi}{4n}\right)\xrightarrow[n\to\infty]{}\frac4\pi\int_0^{\pi/4}\tan^2x\,dx.$

    3. Use $\tan^2x=\sec^2x-1$: $\int_0^{\pi/4}\tan^2x\,dx=\Big[\tan x-x\Big]_0^{\pi/4}=\left(1-\frac\pi4\right)-0=1-\frac\pi4.$ …

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