MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{{{\sin }^{ - 1}}x}} = $
- A$2$
- ✓$1$
- C$-1$
- DNone of these
So $\mathop {\lim }\limits_{y \to 0} \frac{{\sqrt {1 + \sin y} - \sqrt {1 - \sin y} }}{y}$
$(\because \,\,\,x \to 0 \Rightarrow y \to 0)$
$( $ Now multiply it by $\frac{\sqrt{1+\sin y}+\sqrt{1-\sin y}}{\sqrt{1+\sin y}+\sqrt{1-\sin y}}$ and solve $) $
$= 1$
Aliter : Apply $ L-$ Hospital’s rule.
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$\lim _{t \rightarrow x} \frac{f(x) \sin t-f(t) \sin x}{t-x}=\sin ^2 x \text { for all } x \in(0, \pi)$
If $f \left(\frac{\pi}{6}\right)=-\frac{\pi}{12}$, then which of the following statement(s) is (are) TRUE?
$(A)$ $f \left(\frac{\pi}{4}\right)=\frac{\pi}{4 \sqrt{2}}$
$(B)$ $f(x)<\frac{x^4}{6}-x^2$ for all $x \in(0, \pi)$
$(C)$ There exists $\alpha \in(0, \pi)$ such that $f ^{\prime}(\alpha)=0$
$(D)$ $f ^{\prime \prime}\left(\frac{\pi}{2}\right)+ f \left(\frac{\pi}{2}\right)=0$