Q.Compute $\lim_{x\to 0} \left(\frac{e^x - 1}{\sqrt{1+x} - 1}\right)$.

Showing the 12 most recent of 31 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.$\displaystyle\lim_{x \to 0} \dfrac{\sin 4x}{\sin 2x} = ?$

   (a) 0

   (b) 2

   (c) $\dfrac{1}{2}$

   (d) 1

   ›Reveal solutionSolution

   Rewrite using the standard limit $\lim_{\theta\to0}\dfrac{\sin\theta}{\theta}=1$ applied separately to $4x$ and $2x$.

   $\lim_{x\to0}\dfrac{\sin4x}{\sin2x} = \lim_{x\to0}\dfrac{\dfrac{\sin4x}{4x}\times 4x}{\dfrac{\sin2x}{2x}\times 2x} = \lim_{x\to0}\dfrac{\sin4x}{4x}\times\dfrac{2x}{\sin2x}\times\dfrac{4x}{2x}$ …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.The value of $\displaystyle\lim_{x \to 0} \dfrac{\tan x}{x}$ is

   (a) $0$

   (b) $1$

   (c) $-1$

   (d) $2$

   ›Reveal solutionSolution

   $\lim_{x\to0}\dfrac{\tan x}{x}=1$, using the standard limit $\lim_{x\to0}\dfrac{\sin x}{x}=1$ together with $\cos 0=1$.

   Write $\tan x = \dfrac{\sin x}{\cos x}$, so

   $\dfrac{\tan x}{x} = \dfrac{\sin x}{x}\cdot\dfrac{1}{\cos x}$

   …

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3. CBSE 2026Set ANNUAL1 mark
   Q.Write value of $\displaystyle\lim_{x\to 0} \dfrac{\sin x}{x}$.

   ›Reveal solutionSolution

   $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$ is a standard limit, provable via the sandwich (squeeze) theorem using the unit circle.

   For small $x$ (in radians), geometric comparison of the areas of a triangle, a circular sector, and a larger triangle bounding the unit circle gives $\sin x < x < \tan x$ for $0<x<\tfrac{\pi}{2}$.

   …

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4. CBSE 2025Set ANNUAL1 markMCQ
   Q.$\lim_{x \to 0} \dfrac{\sin x}{x} =$

   (a) 0

   (b) 1

   (c) -1

   (d) $1/2$

   ›Reveal solutionSolution

   $\lim_{x\to0}\dfrac{\sin x}{x}=1$.

   This is one of the standard trigonometric limits proved geometrically (comparing areas of triangles and a sector in a unit circle) in the NCERT text. It is used repeatedly to differenti …

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5. CBSE 2025Set ANNUAL1 markMCQ
   Q.$\displaystyle\lim_{x \to 0}\left(\dfrac{\sin ax}{\sin bx}\right)$ is equal to

   (a) $\dfrac{b}{a}$

   (b) $\dfrac{a}{b}$

   (c) $\dfrac{-a}{b}$

   (d) $\dfrac{-b}{a}$

   ›Reveal solutionSolution

   As $x\to0$, $\sin(ax)\approx ax$ and $\sin(bx)\approx bx$, so the ratio tends to $a/b$.

   $\lim_{x\to0}\frac{\sin ax}{\sin bx} = \lim_{x\to0}\frac{\dfrac{\sin ax}{ax}\cdot ax}{\dfrac{\sin bx}{bx}\cdot bx} = \lim_{x\to0}\frac{\dfrac{\sin ax}{ax}}{\dfrac{\sin bx}{bx}}\cdot\frac{a}{b}$

   …

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6. CBSE 2025Set ANNUAL1 markMCQ
   Q.$\displaystyle\lim_{x \to 0}\left(\dfrac{3^{2x} - 1}{2^{3x} - 1}\right)$ is equal to

   (a) $\dfrac{\log 9}{\log 8}$

   (b) $\dfrac{\log 8}{\log 9}$

   (c) $\dfrac{\log 2}{\log 3}$

   (d) $\dfrac{\log 3}{\log 2}$

   ›Reveal solutionSolution

   Rewriting $3^{2x}=9^x$ and $2^{3x}=8^x$ and applying the standard limit $\lim_{x\to0}\frac{k^x-1}{x}=\log k$ to numerator and denominator gives $\dfrac{\log9}{\log8}$.

   $\lim_{x\to0}\frac{3^{2x}-1}{2^{3x}-1} = \lim_{x\to0}\frac{9^x-1}{8^x-1}$

   Divide numerator and denominator by $x$:

   $= \lim_{x\to0}\frac{\dfrac{9^x-1}{x}}{\dfrac{8^x-1}{x}}$

   …

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7. CBSE 2025Set ANNUAL1 mark
   Q.Write the value of $\displaystyle\lim_{x \to 0} \dfrac{\tan x}{x}$.

   ›Reveal solutionSolution

   The standard limit $\displaystyle\lim_{x \to 0}\dfrac{\tan x}{x} = 1$ follows from $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$ and $\displaystyle\lim_{x\to 0}\cos x = 1$.

   $\dfrac{\tan x}{x} = \dfrac{\sin x}{x} \cdot \dfrac{1}{\cos x}$.

   …

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8. CBSE 2024Set ANNUAL1 markMCQ
   Q.$\displaystyle\lim_{x \to \pi} \dfrac{\sin(\pi-x)}{\pi(\pi-x)} =$

   (a) $\dfrac{1}{\pi}$

   (b) $\dfrac{1}{\pi^2}$

   (c) 1

   (d) None of these

   ›Reveal solutionSolution

   Substitute $t = \pi - x$ to turn this into the standard limit $\lim_{t\to0} \dfrac{\sin t}{t} = 1$.

   Let $t = \pi - x$. As $x \to \pi$, $t \to 0$. Rewrite the limit:

   $\lim_{x\to\pi} \frac{\sin(\pi-x)}{\pi(\pi-x)} = \lim_{t\to0} \frac{\sin t}{\pi t} = \frac{1}{\pi}\lim_{t\to0}\frac{\sin t}{t}$

   …

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9. CBSE 2024Set ANNUAL1 mark
   Q.Find: lim(x→0) (a^x - 1) / x = ?

   ›Reveal solutionSolution

   This is a standard limit result: $\lim_{x\to0}\frac{a^x-1}{x}=\ln a$ (natural log of $a$).

   Write $a^x = e^{x\ln a}$. Then

   $\dfrac{a^x-1}{x} = \dfrac{e^{x\ln a}-1}{x} = \ln a\cdot\dfrac{e^{x\ln a}-1}{x\ln a}$.

   …

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10. CBSE 2024Set ANNUAL1 markMCQ
    Q.$\displaystyle\lim_{x\to 0}\dfrac{a^x-b^x}{x}=$

    (a) $\log\left(\dfrac{b}{a}\right)$

    (b) $\log ab$

    (c) $\dfrac{a}{b}$

    (d) $\log\left(\dfrac{a}{b}\right)$

    ›Reveal solutionSolution

    The limit equals $\log\left(\dfrac{a}{b}\right)$.

    Write the limit as a difference of two standard limits:

    $\lim_{x\to0}\dfrac{a^x-b^x}{x}=\lim_{x\to0}\dfrac{a^x-1}{x}-\lim_{x\to0}\dfrac{b^x-1}{x}=\log a-\log b=\log\left(\dfrac{a}{b}\right),$ …

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11. CBSE 2024Set ANNUAL1 markMCQ
    Q.The value of $\displaystyle\lim_{x\to 0}\dfrac{e^x-1}{x} = \ldots\ldots$

    (a) $0$

    (b) $1$

    (c) $\infty$

    (d) $e^x$.

    ›Reveal solutionSolution

    $\displaystyle\lim_{x\to0}\dfrac{e^x-1}{x}=1$.

    Using the series expansion $e^x = 1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots$, we get

    $e^x-1 = x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots,$

    so

    $\dfrac{e^x-1}{x}=1+\dfrac{x}{2!}+\dfrac{x^2}{3!}+\cdots.$ …

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12. CBSE 2024Set ANNUAL1 markMCQ
    Q.The value of $\displaystyle\lim_{x\to 0}\dfrac{\tan x}{x} = \ldots\ldots$

    (a) $x$

    (b) $1$

    (c) $0$

    (d) $\infty$.

    ›Reveal solutionSolution

    $\displaystyle\lim_{x\to0}\dfrac{\tan x}{x}=1$.

    Write $\tan x = \dfrac{\sin x}{\cos x}$, so

    $\dfrac{\tan x}{x}=\dfrac{\sin x}{x}\cdot\dfrac{1}{\cos x}.$ …

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