lim _{x rightarrow 0^{-}} frac{sqrt{frac{1}{2 | vedclass

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Question

(B) આપેલ લક્ષ $L = \lim _{x \rightarrow 0^{-}} \frac{\sqrt{\frac{1}{2}(1-\cos ^2 x)}}{x}$ છે.નિત્યસમ $1-\cos ^2 x = \sin ^2 x$ નો ઉપયોગ કરતા:$L = \lim

Vedclass · Accepted Answer

$-\frac{1}{\sqrt{2}}$

Vedclass · Answer

(B) આપેલ લક્ષ $L = \lim _{x ightarrow 0^{-}} \frac{\sqrt{\frac{1}{2}(1-\cos ^2 x)}}{x}$ છે.નિત્યસમ $1-\cos ^2 x = \sin ^2 x$ નો ઉપયોગ કરતા:$L = \lim _{x ightarrow 0^{-}} \frac{\sqrt{\frac{1}{2} \sin ^2 x}}{x} = \lim _{x ightarrow 0^{-}} \frac{\frac{1}{\sqrt{2}} |\sin x|}{x}$.અહીં $x ightarrow 0^{-}$,તેથી $x $L = \frac{1}{\sqrt{2}} \lim _{x ightarrow 0^{-}} \frac{-\sin x}{x} = -\frac{1}{\sqrt{2}} \lim _{x ightarrow 0^{-}} \frac{\sin x}{x}$.આપણે જાણીએ છીએ કે $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$,તેથી $L = -\frac{1}{\sqrt{2}} 	imes 1 = -\frac{1}{\sqrt{2}}$.

$\lim _{x \rightarrow 0^{-}} \frac{\sqrt{\frac{1}{2}(1-\cos ^2 x)}}{x}$ ની કિંમત શોધો.

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