1. **State the problem:** Find the limit $$\lim_{x \to \frac{\pi}{\gamma}^+} \frac{1}{\cos x}$$ where $x$ approaches $\frac{\pi}{\gamma}$ from the right. 2. **Recall the behavior of cosine:** The cosine function, $\cos x$, is continuous and periodic with period $2\pi$. It equals zero at $x = \frac{\pi}{2} + k\pi$ for integers $k$. 3. **Analyze the denominator near $x = \frac{\pi}{\gamma}$:** Since the limit is from the right, consider values $x > \frac{\pi}{\gamma}$. The value of $\cos x$ near $\frac{\pi}{\gamma}$ depends on $\gamma$. 4. **Key observation:** If $\frac{\pi}{\gamma}$ is a point where $\cos x = 0$, then the denominator approaches zero, and the fraction may tend to $\pm \infty$ or not exist. 5. **Check if $\cos \left( \frac{\pi}{\gamma} \right) = 0$:** This happens if $\frac{\pi}{\gamma} = \frac{\pi}{2} + k\pi$ for some integer $k$. 6. **Solve for $\gamma$ if zero denominator:** $$\frac{\pi}{\gamma} = \frac{\pi}{2} + k\pi \implies \frac{1}{\gamma} = \frac{1}{2} + k \implies \gamma = \frac{1}{\frac{1}{2} + k}$$ 7. **Assuming $\cos \left( \frac{\pi}{\gamma} \right) \neq 0$, the limit is simply:** $$\lim_{x \to \frac{\pi}{\gamma}^+} \frac{1}{\cos x} = \frac{1}{\cos \left( \frac{\pi}{\gamma} \right)}$$ 8. **If $\cos \left( \frac{\pi}{\gamma} \right) = 0$, analyze the sign of $\cos x$ near $\frac{\pi}{\gamma}$ from the right:** - Since cosine crosses zero, near $x = \frac{\pi}{\gamma}$ from the right, $\cos x$ changes sign. - The limit tends to $+\infty$ or $-\infty$ depending on the sign of $\cos x$ just to the right. **Final answer:** - If $\cos \left( \frac{\pi}{\gamma} \right) \neq 0$, then $$\lim_{x \to \frac{\pi}{\gamma}^+} \frac{1}{\cos x} = \frac{1}{\cos \left( \frac{\pi}{\gamma} \right)}$$ - If $\cos \left( \frac{\pi}{\gamma} \right) = 0$, the limit does not exist (tends to infinity with sign depending on $\gamma$).