17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.


lim_x→0 csc 6x sin 7x

Verified step by step guidance

1. First, recognize that the limit involves the product of two trigonometric functions: \( \csc(6x) \) and \( \sin(7x) \). Rewrite \( \csc(6x) \) as \( \frac{1}{\sin(6x)} \). The expression becomes \( \lim_{x \to 0} \frac{\sin(7x)}{\sin(6x)} \).
2. Check the form of the limit as \( x \to 0 \). Both the numerator \( \sin(7x) \) and the denominator \( \sin(6x) \) approach 0, resulting in an indeterminate form \( \frac{0}{0} \). This suggests that l'Hôpital's Rule might be applicable.
3. Apply l'Hôpital's Rule, which states that if the limit \( \lim_{x \to a} \frac{f(x)}{g(x)} \) results in an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \), provided the latter limit exists.
4. Differentiate the numerator and the denominator separately. The derivative of \( \sin(7x) \) is \( 7\cos(7x) \), and the derivative of \( \sin(6x) \) is \( 6\cos(6x) \). Substitute these derivatives back into the limit: \( \lim_{x \to 0} \frac{7\cos(7x)}{6\cos(6x)} \).
5. Evaluate the new limit as \( x \to 0 \). Since \( \cos(7x) \to \cos(0) = 1 \) and \( \cos(6x) \to \cos(0) = 1 \), the limit simplifies to \( \frac{7}{6} \). Thus, the original limit evaluates to \( \frac{7}{6} \).