Evaluate the limit using L'Hopital's rule $$ \lim _{x \rightarrow \frac{\pi}{2}} 5 \cos (7 x) \sec (5 x) $$

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To evaluate the limit $$ \lim _{x \rightarrow \frac{\pi}{2}} 5 \cos (7 x) \sec (5 x) $$, we first rewrite the expression in a more manageable form:

$$
\sec(5x) = \frac{1}{\cos(5x)}
$$

Thus, the limit can be rewritten as:

$$
\lim _{x \rightarrow \frac{\pi}{2}} 5 \cos(7x) \cdot \frac{1}{\cos(5x)} = \lim _{x \rightarrow \frac{\pi}{2}} \frac{5 \cos(7x)}{\cos(5x)}
$$

Now, we need to evaluate this limit as $$ x $$ approaches $$ \frac{\pi}{2} $$. At $$ x = \frac{\pi}{2} $$:

- $$ \cos(7x) = \cos\left(7 \cdot \frac{\pi}{2}\right) = \cos\left(\frac{7\pi}{2}\right) = 0 $$
- $$ \cos(5x) = \cos\left(5 \cdot \frac{\pi}{2}\right) = \cos\left(\frac{5\pi}{2}\right) = 0 $$

This results in the indeterminate form $$ \frac{0}{0} $$, which allows us to apply L'Hôpital's rule. We will differentiate the numerator and the denominator:

1. Differentiate the numerator: $$ 5 \cos(7x) $$
2. Differentiate the denominator: $$ \cos(5x) $$

Let's compute the derivatives:

- The derivative of $$ 5 \cos(7x) $$ is $$ -35 \sin(7x) $$
- The derivative of $$ \cos(5x) $$ is $$ -5 \sin(5x) $$

Now we can apply L'Hôpital's rule:

$$
\lim _{x \rightarrow \frac{\pi}{2}} \frac{5 \cos(7x)}{\cos(5x)} = \lim _{x \rightarrow \frac{\pi}{2}} \frac{-35 \sin(7x)}{-5 \sin(5x)} = \lim _{x \rightarrow \frac{\pi}{2}} \frac{7 \sin(7x)}{\sin(5x)}
$$

Now we will evaluate this limit as $$ x $$ approaches $$ \frac{\pi}{2} $$:

- $$ \sin(7x) = \sin\left(7 \cdot \frac{\pi}{2}\right) = \sin\left(\frac{7\pi}{2}\right) = 1 $$
- $$ \sin(5x) = \sin\left(5 \cdot \frac{\pi}{2}\right) = \sin\left(\frac{5\pi}{2}\right) = 1 $$

Thus, we have:

$$
\lim _{x \rightarrow \frac{\pi}{2}} \frac{7 \sin(7x)}{\sin(5x)} = \frac{7 \cdot 1}{1} = 7
$$

Therefore, the final result is:

$$
\lim _{x \rightarrow \frac{\pi}{2}} 5 \cos(7x) \sec(5x) = 7
$$
The limit is 7.

Evaluate the limit using L’Hopital’s rule

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