# Chapter 7 - Exponential Functions - 7.7 L'Hôpital's Rule - Exercises - Page 366: 9

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#### Work Step by Step

Since we have $$\lim _{x \rightarrow 0} \frac{\cos 2x-1}{\sin 5x }=\frac{0}{0}$$ then we can apply L’Hôpital’s Rule as follows $$\lim _{x \rightarrow 0} \frac{-2\sin 2x}{5\cos 5x }=\frac{0}{5}=0.$$

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