Q.$\displaystyle\lim_{x\to0}\dfrac{\sqrt{1-\cos2x}}{x}$

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 $2x$.

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

   As $x\to0$, $\dfrac{\sin4x}{4x}\to1$ and $\dfrac{\sin2x}{2x}\to1$, leaving:

   $= 1\times1\times\dfrac{4x}{2x} = 2$

   ✓Final answer

   (b) 2.

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

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

   As $x\to 0$: $\dfrac{\sin x}{x}\to 1$ (the standard trigonometric limit) and $\cos x \to \cos 0 = 1$, so $\dfrac{1}{\cos x}\to 1$.

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

   ✓Final answer

   $\displaystyle\lim_{x\to 0}\dfrac{\tan x}{x}=1$, which is option (b).

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 $0<x<\tfrac{\pi}{2}$.

   Dividing through by $\sin x$ and taking reciprocals: $\cos x < \dfrac{\sin x}{x} < 1$.

   As $x\to0$, $\cos x\to1$, so by the squeeze theorem $\dfrac{\sin x}{x}\to1$.

   ✓Final answer

   $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$.

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 differentiate $\sin x$, $\cos x$, $\tan x$, etc.

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

   ✓Final answer

   (b) 1.

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 $a/b$.

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

   Since $\lim_{x\to0}\dfrac{\sin ax}{ax} = 1$ and $\lim_{x\to0}\dfrac{\sin bx}{bx}=1$ (standard limit $\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$, applied with $\theta=ax$ and $\theta=bx$ respectively):

   $\lim_{x\to0}\frac{\sin ax}{\sin bx} = 1 \times \frac{a}{b} = \frac{a}{b}$

   ✓Final answer

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

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 $x$:

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

   Using the standard limit $\lim_{x\to0}\dfrac{a^x-1}{x} = \log a$ (natural log, or any consistent base):

   $= \frac{\log 9}{\log 8}$

   ✓Final answer

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

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 $\displaystyle\lim_{x\to 0}\cos x = 1$.

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

   As $x \to 0$: $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x} = 1$ (standard result) and $\displaystyle\lim_{x\to 0}\dfrac{1}{\cos x} = \dfrac{1}{1} = 1$.

   So $\displaystyle\lim_{x\to 0}\dfrac{\tan x}{x} = 1 \times 1 = 1$.

   ✓Final answer

   $\displaystyle\lim_{x \to 0} \dfrac{\tan x}{x} = 1$.

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:

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

   Using the standard trigonometric limit $\displaystyle\lim_{t\to0}\frac{\sin t}{t} = 1$:

   $= \frac{1}{\pi} \times 1 = \frac{1}{\pi}$

   ✓Final answer

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

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

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

   As $x\to 0$, let $t=x\ln a\to 0$, and using the standard limit $\lim_{t\to0}\frac{e^t-1}{t}=1$:

   $\lim_{x\to0}\dfrac{a^x-1}{x} = \ln a \times 1 = \ln a$.

   ✓Final answer

   $\displaystyle\lim_{x\to0}\frac{a^x-1}{x} = \ln a$.

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:

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

    using the standard result $\displaystyle\lim_{x\to0}\dfrac{a^x-1}{x}=\log a$.

    ✓Final answer

    $\displaystyle\lim_{x\to0}\dfrac{a^x-b^x}{x}=\log\left(\dfrac{a}{b}\right)$ — option (d).

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

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

    As $x\to0$, every term after the $1$ vanishes, giving the limit $1$. This is a standard result used repeatedly in calculus.

    ✓Final answer

    The correct option is (b) $1$.

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

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

    Using the standard limit $\displaystyle\lim_{x\to0}\dfrac{\sin x}{x}=1$ and $\displaystyle\lim_{x\to0}\cos x=1$:

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

    ✓Final answer

    The correct option is (b) $1$.