_ x 0 x^3 x 1 - x = - Vidyadip

MCQ

$\mathop {\lim }\limits_{x \to 0} \frac{{{x^3}\cot x}}{{1 - \cos x}} = $

A
$0$
B
$1$
✓
$2$
D
$-2$

✓

Answer

Correct option: C.

$2$

c
(c) $\mathop {\lim }\limits_{x \to 0} \,\,\frac{{{x^3}\cot x}}{{1 - \cos x}} = \mathop {\lim }\limits_{x \to 0} \,\left( {\frac{{{x^3}\cot x}}{{1 - \cos x}} \times \frac{{1 + \cos x}}{{1 + \cos x}}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} \,{\left( {\frac{x}{{\sin x}}} \right)^3} \times \mathop {\lim }\limits_{x \to 0} \,\cos x \times \mathop {\lim }\limits_{x \to 0} \,(1 + \cos x) = 2$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from 12. limits→12. limits — all question types→MATHS STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The equation of the line through the points (1, 5) and (2, 3) is:

Let $\log _a b=4, \log _c d=2$, where $a, b, c, d$ are natural numbers. Given that $b-d=7$, the value of $c-a$ is

If $\mathop {\lim }\limits_{x \to 2} \frac{{a{x^2} + bx + c}}{{{{(2x - 1)}^2}}} = \frac{1}{2}$ then $\mathop {\lim }\limits_{x \to 2} \frac{{(x - a)(x - b)(x - c)}}{{x - 2}}$ is

Solve $\text{x}^2 + 1 = 0$.

The equation formed by decreasing each root of $a{x^2} + bx + c = 0$ by $1$ is $2{x^2} + 8x + 2 = 0,$ then

If straight lines $ax + by + p = 0$ and $x\cos \alpha + y\sin \alpha - p = 0$ include an angle $\pi /4$ between them and meet the straight line $x\sin \alpha - y\cos \alpha = 0$ in the same point, then the value of ${a^2} + {b^2}$ is equal to

Three randomly chosen nonnegative integers $x, y$ and $z$ are found to satisfy the equation $x+y+z=10$. Then the probability that $z$ is even, is

For any odd integer $n \ge 1$, ${n^3} - {(n - 1)^3} + ........... + {( - 1)^{n - 1}}{1^3} = $

Let $\theta, \phi \in[0,2 \pi]$ be such that $2 \cos \theta(1-\sin \phi)=\sin ^2 \theta\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \phi-1, \tan (2 \pi-\theta)>0$ and $-1 < \sin \theta < -\frac{\sqrt{3}}{2}$. Then $\phi$ cannot satisfy

$(A)$ $0 < \phi<\frac{\pi}{2}$ $(B)$ $\frac{\pi}{2} < \phi<\frac{4 \pi}{3}$

$(C)$ $\frac{4 \pi}{3} < \phi<\frac{3 \pi}{2}$ $(D)$ $\frac{3 \pi}{2} < \phi < 2 \pi$

Let $L$ be a normal to the parabola $y^2=4 x$. If $L$ passes through the point $(9,6)$, then $L$ is given by

$(A)$ $y-x+3=0$ $(B)$ $y+3 x-33=0$ $(C)$ $y+x-15=0$ $(D)$ $y-2 x+12=0$