_ x 0 | x| x is equal to - Vidyadip

MCQ

$\lim _{x \rightarrow 0} \frac{|\sin x|}{x}$ is equal to

A
$0$
B
Does not exist
C
1
D
-1

✓

Answer

(b) Does not exist
Explanation: Given, $\lim _{x \rightarrow 0} \frac{|\sin x|}{x}$
$\begin{array}{l} LHL =\lim _{x \rightarrow 0^{-}} \frac{-\sin x}{x}=-1 \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \\ RHL =\lim _{x \rightarrow 0^{+}} \frac{\sin x}{x}=1\end{array}$
LHL $\neq$ RHL, So the limit does not exist.

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 "M.C.Q (1 Marks)" from Model Paper 9→Model Paper 9 — all question types→MATHS STD 11 Science question papers→Rajasthan Board STD 11 Science question papers→Rajasthan Board question paper generator→

Similar questions

If $\cot\text{x}-\tan\text{x}=\sec\text{x},$ then, x is equal to:

$2\text{n}\pi+\frac{3\pi}{2},\text{n}\in\text{Z}$
$\text{n}\pi+(-1)^\text{n}\frac{\pi}{6},\text{n}\in\text{Z}$
$\text{n}\pi+\frac{\pi}{2},\text{n}\in\text{Z}$
None of these.

If r^th term is the middle term in the expansion of $\Big(\text{x}^{2}-\frac{1}{2\text{x}}\Big)^{20},$ then (r + 3)^th term is:

${^\text{20}}\text{C}_{\text{14}}\ \Big(\frac{\text{x}}{2^{14}}\Big)$
${^\text{20}}\text{C}_{\text{12}}\ \text{x}^{2}\ 2^{-12}$
$-{^\text{20}}\text{C}_{\text{7}}\ \text{x}\ 2^{-13}$
None of these.

If ${ }^{n-1} C _3+{ }^{n-1} C _4>{ }^n C _3$ then the value of $x$ is:

If (-3, 2) lies on the circle x² + y² + 2gx + 2fy + c = 0 which is concentric with the circle x² + y² + 6x + 8y - 5 = 0, then c =

$\text{If}\ \sin2\theta+\sin2\phi=\frac{1}{2}\ \text{and}\ \cos2\theta\\ \ +\cos2\phi=\frac{3}{2},\text{then}\cos^2(\theta-\phi)=$

$\frac{3}{8}$
$\frac{5}{8}$
$\frac{3}{4}$
$\frac{5}{4}$

If ${^\text{n+1}}\text{C}_{\text{3}}=2.{^\text{n}}\text{C}_{\text{2}},$ then n:

The ________ is the difference between the greatest and the least value of the variate:

$\frac{\sin3\text{x}}{1+2\cos2\text{x}}$ is equal to:

If in the binomial expansion of (1 + x)n where n is a natural number, the coefficients of the 5th, 6th and 7th terms are in A.P., then n is equal to:

If $(2^\text{n}+1)\text{x}=\pi,$ then $2^\text{n}\cos\text{x}\cos2\text{x}^2\text{x}\cos^\text{n-1}\text{x}=$