- A$\sqrt 2 $
- B-$\;\sqrt 2 $
- C$\frac{1}{{\sqrt 2 }}$
- ✓does not exist
$ = \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt 2 |\sin (x - 2)|}}{{x - 2}}$
$R.H.L. = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{\sqrt 2 \sin (x - 2)}}{{(x - 2)}} = - \sqrt 2 $
$R.H.L. = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt 2 \sin (x - 2)}}{{(x - 2)}} = - \sqrt 2 $
Thus $L . H . L . \neq R . H . L$
Hence, $\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {1 - \cos \{ 2(x - 2)\} } }}{{x - 2}}$ does not exist.
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Statement $-1$ : $R$ is symmetric
Statement $-2$ : $R$ is reflexive
Statement $-3$ : $R$ is transitive, then thecorrect sequence of given statements is
(where $T$ means true and $F$ means false)
$x^5 - 40x^4 + px^3 + qx^2 + rx + s = 0$ are in $G.P.$ The sum of their reciprocals is $10$. Then the value of $\left| s \right|$ is