MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{\tan x - \sin x}}{{{x^3}}} = $
- ✓$\frac{1}{2}$
- B$ - \frac{1}{2}$
- C$\frac{2}{3}$
- Dએકપણ નહી.
$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{\sin x\,\left( {2\,\,{{\sin }^2}\frac{x}{2}} \right)}}{{{x^3}\,\cos x}} $
$= \mathop {\lim }\limits_{x \to 0} \,\left[ {\frac{{\sin x}}{x}.\frac{2}{{\cos x}}.\frac{{{{\sin }^2}\frac{x}{2}}}{{{{\left( {\frac{x}{2}} \right)}^2}}}.\frac{1}{4}} \right] = \frac{1}{2}$.
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$A=\{m \in R:$ : સમીકરણ $x^{2}-(m+1) x+m+4=0$ ના બંને બીજો વાસ્તવિક છે $\}$ અને $B=[-3,5)$
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