Question
$\mathop {\lim }\limits_{x \to 1} \frac{{1 - \sqrt x }}{{{{({{\cos }^{ - 1}}x)}^2}}} = $
$\mathop {\lim }\limits_{x \to 1} \,\frac{{1 - \sqrt x }}{{{{({{\cos }^{ - 1}}x)}^2}}} = \mathop {\lim }\limits_{y \to 0} \,\frac{{1 - \sqrt {\cos y} }}{{{y^2}}}$
अब इसका परिमेयीकरण करने पर,
$\mathop {\lim }\limits_{y \to 0} \,\frac{{(1 - \cos y)}}{{{y^2}(1 + \sqrt {\cos y} )}}$
$ = \mathop {\lim }\limits_{y \to 0} \,\frac{{1 - \cos y}}{{{y^2}}}\,.\,\mathop {\lim }\limits_{y \to 0} \,\frac{1}{{1 + \sqrt {\cos y} }} $
$= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.$
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$\mathrm{S}_{\mathrm{k}}(\mathrm{x})=\mathrm{C}_{\mathrm{k}} \mathrm{x}+\mathrm{k} \int_0^{\mathrm{x}} \mathrm{S}_{\mathrm{k}-1}(\mathrm{t}) \mathrm{dt}$, हैं, जहाँ
$\mathrm{C}_0=1, \mathrm{C}_{\mathrm{k}}=1-\int_0^{\mathrm{l}} \mathrm{S}_{\mathrm{k}-1}(\mathrm{x}) \mathrm{dx}, \mathrm{k}=1,2,3 \ldots$ हैं।
तो $\mathrm{S}_2(3)+6 \mathrm{C}_3$ बराबर है