To find the limit as \( x \) approaches \( 0^+ \) of the expression \(\frac{5}{x} - \frac{7}{\sin x}\), we need to analyze the behavior of each term separately as \( x \) approaches zero from the positive side. The term \(\frac{5}{x}\) tends to infinity, while \(\frac{7}{\sin x}\) also tends to infinity, but at a different rate. We can use the Taylor series expansion for \(\sin x\) around zero, which is \(\sin x \approx x\), to simplify the expression and find the limit.
We need to evaluate the limit
\[
\lim_{x \rightarrow 0^{+}} \left( \frac{5}{x} - \frac{7}{\sin x} \right).
\]
As \( x \) approaches \( 0^{+} \), both terms \( \frac{5}{x} \) and \( \frac{7}{\sin x} \) approach infinity. Therefore, we need to analyze their rates of growth.
Using the Taylor series expansion for \( \sin x \) around \( x = 0 \):
\[
\sin x \approx x \quad \text{as } x \to 0,
\]
we can rewrite the limit as:
\[
\lim_{x \rightarrow 0^{+}} \left( \frac{5}{x} - \frac{7}{x} \right) = \lim_{x \rightarrow 0^{+}} \left( \frac{5 - 7}{x} \right) = \lim_{x \rightarrow 0^{+}} \left( \frac{-2}{x} \right).
\]
As \( x \) approaches \( 0^{+} \), \( \frac{-2}{x} \) approaches \( -\infty \). Therefore, we conclude that:
\[
\lim_{x \rightarrow 0^{+}} \left( \frac{5}{x} - \frac{7}{\sin x} \right) = -\infty.
\]
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