MCQ
$\lim _{x \rightarrow 0} \cos \frac{1}{x}$
- Ais continuous at $x=0$
- Bis differentiable at $x=0$
- ✓does not exist
- Dnone of these
✓
Answer
Correct option: C.
does not exist
(C)
$\lim _{x \rightarrow 0} \cos \frac{1}{x}$ oscillates between -1 and 1 .
∴ Limit doesn't exist.
$\lim _{x \rightarrow 0} \cos \frac{1}{x}$ oscillates between -1 and 1 .
∴ Limit doesn't exist.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The point of contact of the tangent $y=x+2$ to the hyperbola $5 x^2-9 y^2=45$ is
→If $\text{R}=\{(\text{x, y}):\text{x, y}\in\text{Z},\text{ x}^2+\text{y}^2\leq4\}$ is a relation on $Z,$ then the domain of $R$ is:
→$\text {If } f(x)=\left\{\begin{array}{ll}\frac{1}{2} \sin x^2, & x \neq 0 \\0, & x=0\end{array}\right. \text {, then }$
→If $A , B$ and C are the angles of a plain triangle and $\tan \frac{A}{2}=\frac{1}{3}, \tan \frac{B}{2}=\frac{2}{3}$, then $\tan \frac{C}{2}$ is equal to
→If $\tan\Big(\frac{\pi}{4}+\text{x}\Big)+\tan\Big(\frac{\pi}{4}-\text{x}\Big)=\lambda\sec2\text{x},$ then :
→→
The range of $\text{f(x)}=\cos[\text{x}],$ for $-\frac{\pi}{2}<\text{x}<\frac{\pi}{2}$ is :
→If O is the origin and Q is a variable point on $x^2=4 y$, then locus of the midpoint of seg OQ is
→The point A is (3, 0) and abscissa of the point B is -3. A variable point P is such that ordinates of P and B are equal, then equation of locus of P such that AP = PB is
If $n = 1,2,3, ...,$ then $\cos\alpha\cos4\alpha\ ...\cos2^{\text{n}-1}\alpha$ is equal to :
→