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Compute the limit of the given function

Evaluate the limit.

    \[ \lim_{x \to 0^+} \frac{1}{\sqrt{x}} \left( \frac{1}{\sin x} - \frac{1}{x} \right). \]


First,

    \[ \frac{1}{\sqrt{x}} \left( \frac{1}{\sin x} - \frac{1}{x} \right) = \frac{1}{\sqrt{x}} \cdot \frac{x - \sin x}{x \sin x} = \frac{x - \sin x}{x^{\frac{3}{2}} \sin x}. \]

Then, we multiply inside the limit by \frac{\sin x}{x} since \lim_{x \to 0} \frac{\sin x}{x} = 1,

    \begin{align*}  \lim_{x \to 0^+} \frac{1}{\sqrt{x}} \left(\frac{1}{\sin x} - \frac{1}{x} \right) &= \lim_{x \to 0^+} \frac{x - \sin x}{x^{\frac{3}{2}} \sin x} \\[9pt]  &= \lim_{x \to 0^+} \left( \frac{x - \sin x}{x^{\frac{3}{2}} \sin x} \cdot \frac{\sin x}{x} \right) \\[9pt]  &= \lim_{x \to 0^+} \frac{x - \sin x}{x^{\frac{5}{2}}} \\[9pt]  &= \lim_{x \to 0^+} \frac{1 - \cos x}{\frac{5}{2} x^{\frac{3}{2}}} &(\text{L'Hopital's})\\[9pt]  &= \lim_{x \to 0^+} \frac{4 \sin x}{15 \sqrt{x}}  &(\text{L'Hopital's})\\[9pt]  &= \lim_{x \to 0^+} \frac{8 \sqrt{x} \cos x}{15}  &(\text{L'Hopital's})\\[9pt]  &= 0. \end{align*}

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