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\(\frac{\sqrt{\sin(x)} - \sqrt{\sin(x_0)}}{x - x_0}, \quad x \neq x_0\)
\({p}, {x=x_0}\)

Jeżeli \( x_0 \in (0 +2k \pi , \pi + 2k \pi ) \ , \ k \in Z \) to

\( \Lim_{x \to x_0} \frac{\sqrt{\sin(x)} - \sqrt{\sin(x_0)}}{x - x_0} =
\Lim_{x \to x_0} \left(\frac{\sin(x) - \sin(x_0)}{x- x_0} \cdot \frac{1}{\sqrt{\sin(x)} + \sqrt{\sin(x_0)}} \right) = \\
= \Lim_{x \to x_0} \left( \frac{\sin(\frac{x - x_0}{2})}{\frac{x-x_0}{2}} \cdot \cos(\frac{x+x_0}{2}) \cdot \frac{1}{\sqrt{\sin(x)} + \sqrt{\sin(x_0)}} \right) = \frac{\cos(x_0)}{2\sqrt{\sin(x_0)}}\)
Edit:
Zgodnie z uwagą użytkownika Jerry zastosowałem niepoprawny wzór.
Poniżej wersja oryginalna (błędna):
\( \Lim_{x \to x_0} \frac{\sqrt{\sin(x)} - \sqrt{\sin(x_0)}}{x - x_0} =
\Lim_{x \to x_0} \left(\frac{\sin(x) - \sin(x_0)}{x- x_0} \cdot \frac{1}{\sqrt{\sin(x)} + \sqrt{\sin(x_0)}} \right) = \\
= \Lim_{x \to x_0} \left( \frac{\sin(\frac{x - x_0}{2})}{\frac{x-x_0}{2}} \cdot \cos(\frac{x-x_0}{2}) \cdot \frac{1}{\sqrt{\sin(x)} + \sqrt{\sin(x_0)}} \right) = \frac{1}{2\sqrt{\sin(x_0)}}\)