Q.$\lim_{x\to 0}\dfrac{ax + b}{cx + 1}$

Showing the 12 most recent of 49 on this concept.

1. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.The value of $\lim\limits_{x\to 0}\dfrac{6\sin(x)-2\sin(3x)}{3x^{3}}$ is equal to (A) $\dfrac{-8}{3}$ (B) $\dfrac{8}{3}$ (C) $\dfrac{-8}{9}$ (D) $\dfrac{8}{9}$ (E) $0$

   ›Reveal solutionSolution

   The numerator is a $0/0$ form; expand each sine to third order — the $x$-terms cancel, leaving $8x^3$, and dividing by $3x^3$ gives $\dfrac{8}{3}$.

   Concept. Both numerator and denominator vanish as $x\to0$, so we need the leading non-zero power. The cubic Maclaurin expansion $\sin u=u-\dfrac{u^3}{6}+O(u^5)$ makes the cancellation explicit.

   Step 1 — expand $6\sin x$.

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

   Step 2 — expand $2\sin 3x$.

   $\sin 3x=3x-\frac{(3x)^3}{6}+\cdots=3x-\frac{27x^3}{6}+\cdots=3x-\frac{9x^3}{2}+\cdots,$

   $2\sin 3x=6x-9x^3+\cdots.$

   Step 3 — subtract. …

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2. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.If $\lim_{x \to 3} \dfrac{(2x-k)\tan(x-3)}{x^2 - 6x + 9} = 2$, then the value of 'k' is equal to (A) $2$ (B) $3$ (C) $4$ (D) $6$ (E) $8$

   ›Reveal solutionSolution

   For the limit to be finite the numerator must vanish at $x=3$, forcing $k=6$; then the limit is $2$.

   Since $x^{2}-6x+9=(x-3)^{2}$ and $\tan(x-3)\sim(x-3)$ as $x\to3$:

   $\frac{(2x-k)\tan(x-3)}{(x-3)^{2}}\sim\frac{(2x-k)(x-3)}{(x-3)^{2}}=\frac{2x-k}{x-3}.$ …

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3. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.The value of $\lim_{x \to 0} \frac{sin^{2}x}{1 - cosx}$ is equal to (A) 4 (B) 2 (C) $\frac{1}{2}$ (D) $\frac{1}{4}$ (E) 0

   ›Reveal solutionSolution

   Use $\sin^2x=1-\cos^2x=(1-\cos x)(1+\cos x)$, cancel, giving limit $2$.

   Simplify.

   $\frac{\sin^2x}{1-\cos x}=\frac{(1-\cos x)(1+\cos x)}{1-\cos x}=1+\cos x.$ …

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4. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.The value of $\lim_{x \to 1} \frac{x-1}{3\sqrt{x}-1}$ is equal to (A) 3 (B) $\frac{1}{3}$ (C) 2 (D) $\frac{1}{2}$ (E) 0

   ›Reveal solutionSolution

   The denominator does not vanish at $x=1$ (it equals $2$), so the limit is found by direct substitution: $0/2=0$.

   Concept. A limit is indeterminate only when both numerator and denominator approach $0$ (or both $\infty$). If the denominator has a non-zero limit, the quotient limit is obtained by plugging in directly.

   Step 1 — evaluate numerator and denominator at $x=1$.

   $\text{Numerator: } x-1\ \xrightarrow{x\to1}\ 1-1=0.$

   $\text{Denominator: } 3\sqrt{x}-1\ \xrightarrow{x\to1}\ 3\sqrt{1}-1=3-1=2.$ …

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5. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.The value of $\lim_{x \to 0} \frac{\sqrt{1 - \cos(x^2)}}{1 - \cos x}$ is equal to (A) $\frac{1}{\sqrt{2}}$ (B) $\sqrt{2}$ (C) $\frac{1}{2\sqrt{2}}$ (D) $2\sqrt{2}$ (E) 0

   ›Reveal solutionSolution

   Small-angle: $1-\cos u\approx u^2/2$. Numerator $\sim x^2/\sqrt2$, denominator $\sim x^2/2$, ratio $=\sqrt2$.

   Numerator. $1-\cos(x^2)\approx\dfrac{(x^2)^2}{2}=\dfrac{x^4}{2}$, so $\sqrt{1-\cos(x^2)}\approx\dfrac{x^2}{\sqrt2}$.

   Denominator. $1-\cos x\approx\dfrac{x^2}{2}$.

   Limit. …

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6. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.The value of $\lim_{x \to \frac{\pi}{2}^-} (\tan x - \sec x)$ is equal to (A) $\frac{1}{2}$ (B) $\frac{1}{4}$ (C) $0$ (D) $2$ (E) $1$

   ›Reveal solutionSolution

   Combine to $\frac{\sin x-1}{\cos x}$ and apply L'Hôpital at $x\to\frac{\pi}{2}^-$.

   $\tan x-\sec x=\dfrac{\sin x-1}{\cos x}$, which is $\dfrac{0}{0}$ as $x\to\frac{\pi}{2}$. …

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7. KEAM 2026Set eng-2026-04214 marksMCQ
   Q.The value of $\lim_{x\to 5}\left(\frac{25-x^{2}}{4-\sqrt{x^{2}-9}}\right)$ is (A) $32$ (B) $16$ (C) $8$ (D) $4$ (E) $0$

   ›Reveal solutionSolution

   Rationalizing the denominator cancels $25-x^2$, leaving $4+\sqrt{x^2-9}\to8$.

   Multiply by the conjugate $4+\sqrt{x^2-9}$: …

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8. KEAM 2026Set eng-2026-04214 marksMCQ
   Q.The value of $\lim_{x \to 0} \dfrac{\sqrt{1-\cos 2x}}{|x|}$ is equal to (A) $-2$ (B) $-\sqrt{2}$ (C) $\sqrt{2}$ (D) $1$ (E) $2$

   ›Reveal solutionSolution

   Using $1-\cos2x=2\sin^2x$, the expression becomes $\sqrt2\,\frac{|\sin x|}{|x|}\to\sqrt2$.

   Since $1-\cos2x=2\sin^2x$:

   $\frac{\sqrt{1-\cos2x}}{|x|}=\frac{\sqrt{2\sin^2x}}{|x|}=\sqrt2\,\frac{|\sin x|}{|x|}.$ …

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9. KEAM 2026Set eng-2026-04214 marksMCQ
   Q.The positive integer $n$, such that $\lim_{x \to 3} \dfrac{x^n - 3^n}{x-3} = 108$ (A) $3$ (B) $12$ (C) $6$ (D) $9$ (E) $4$

   ›Reveal solutionSolution

   Apply the standard result $\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=n\,a^{n-1}$; solve $n\cdot 3^{n-1}=108$ to get $n=4$.

   The limit $\lim_{x\to 3}\dfrac{x^n-3^n}{x-3}$ is exactly the derivative of $x^n$ at $x=3$, which equals $n\,x^{n-1}\big|_{x=3}=n\cdot 3^{n-1}$. …

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10. KEAM 2026Set eng-2026-04214 marksMCQ
    Q.Let $\lim_{x \to a} f(x)g(x) = 16$ and $\lim_{x \to a} \dfrac{f(x)}{g(x)} = 4$. If both $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exists, then $\lim_{x \to a} [f(x) + g(x)] =$ (A) $\pm 10$ (B) $-16$ (C) $\pm 2$ (D) $16$ (E) $\pm 4$

    ›Reveal solutionSolution

    Multiply and divide the two given limits to isolate $f$ and $g$: $f=\pm 8$, $g=\pm 2$ (same sign), so $f+g=\pm 10$.

    Let $F=\lim f(x)$, $G=\lim g(x)$. Given $FG=16$ and $F/G=4$.

    Multiplying: $F^2=(FG)(F/G)=16\cdot 4=64\Rightarrow F=\pm 8$.

    Dividing: $G^2=(FG)/(F/G)=16/4=4\Rightarrow G=\pm 2$. …

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11. KEAM 2025Set eng-2025-04234 marksMCQ
    Q.$\lim\limits_{x \to 0} \dfrac{x^2}{\sqrt{2} - \sqrt{1 + \cos x}}$ is equal to (A) $4\sqrt{2}$ (B) $4$ (C) $2\sqrt{2}$ (D) $\sqrt{2}$ (E) $0$

    ›Reveal solutionSolution

    The limit is $4\sqrt2$.

    Concept and Intuition

    Rationalize the denominator, then use $1-\cos x\sim x^2/2$ as $x\to0$.

    Step-by-Step Solution

    1. Multiply by $\dfrac{\sqrt2+\sqrt{1+\cos x}}{\sqrt2+\sqrt{1+\cos x}}$: denominator becomes $2-(1+\cos x)=1-\cos x$.
    2. Expression $=\dfrac{x^2(\sqrt2+\sqrt{1+\cos x})}{1-\cos x}$; as $x\to0$, numerator factor $\to 2\sqrt2$ and $1-\cos x\sim x^2/2$. …

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12. KEAM 2025Set eng-2025-04234 marksMCQ
    Q.$\lim\limits_{x \to 0} \dfrac{\sqrt{\cos^2 x + 3} - \sqrt{\cos^2 x + \sin x + 3}}{x} =$ (A) $\dfrac{1}{4}$ (B) $\dfrac{-1}{4}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{-1}{2}$ (E) $-1$

    ›Reveal solutionSolution

    The limit is $-\dfrac14$.

    Concept and Intuition

    Rationalize the numerator; the difference of the radicands is $-\sin x$, and $\sin x/x\to1$.

    Step-by-Step Solution

    1. Multiply by the conjugate: numerator $\to(\cos^2x+3)-(\cos^2x+\sin x+3)=-\sin x$.
    2. Expression $=\dfrac{-\sin x}{x\left(\sqrt{\cos^2x+3}+\sqrt{\cos^2x+\sin x+3}\right)}$.
    3. As $x\to0$: $\dfrac{-\sin x}{x}\to-1$ and denominator $\to\sqrt4+\sqrt4=4$, so limit $=-\dfrac14$.

    Common Mistakes …

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