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[Dziura-LBN]

\(\displaystyle{ $\lim_{x\to 0}\frac{\sqrt{1-\sqrt{\cos{x}}}}{x}$}\)

Z gory dzieki!

[Calasilyar]

\(\displaystyle{ \lim\limits_{x\to 0} \frac{\sqrt{1-\sqrt{cosx}}}{x}=\lim\limits_{x\to 0} \sqrt{\frac{1-\sqrt{cosx}}{x^{2}}}=\sqrt{\lim\limits_{x\to 0} \frac{1-\sqrt{cosx}}{x^{2}}}=H=
\sqrt{\lim\limits_{x\to 0} \frac{sinx}{4x\sqrt{cosx}}}=\sqrt{\lim\limits_{x\to 0} \frac{sinx}{x}\cdot \frac{1}{4\sqrt{cosx}}}=\sqrt{\frac{1}{4}}=\frac{1}{2}}\)

[max]

\(\displaystyle{ \sqrt{x^{2}} = |x| \not\equiv x}\)

\(\displaystyle{ \lim_{x\to 0}\frac{\sqrt{1 - \sqrt{\cos x}}}{x} = \lim_{x\to 0}\left(\frac{\sqrt{1 - \sqrt{\cos x}}}{x}\cdot \frac{\sqrt{1 + \sqrt{\cos x}}}{\sqrt{1 + \sqrt{\cos x}}}\right) =\\
= \lim_{x\to 0}\frac{\sqrt{1 - \cos x}}{x\sqrt{1 + \sqrt{\cos x}}} =\lim_{x\to 0}\frac{\sqrt{2\sin^{2} \frac{x}{2}}}{x\sqrt{1 + \sqrt{\cos x}}} = \lim_{x\to 0}
\frac{\sqrt{2}|\sin \frac{x}{2}|}{2\cdot \frac{x}{2}\sqrt{1 + \sqrt{\cos x}}}}\)
Stąd:
\(\displaystyle{ \lim_{x\to 0^{+}}\frac{\sqrt{1 - \sqrt{\cos x}}}{x} = \frac{1}{2}\\
\lim_{x\to 0^{-}}\frac{\sqrt{1 - \sqrt{\cos x}}}{x} = -\frac{1}{2}}\)