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


lim_x→0 x csc x

Verified step by step guidance

1. First, recognize that the limit \( \lim_{x \to 0} x \csc x \) can be rewritten using the identity \( \csc x = \frac{1}{\sin x} \). Thus, the expression becomes \( \lim_{x \to 0} \frac{x}{\sin x} \).
2. Observe that as \( x \to 0 \), both the numerator and the denominator approach 0, creating an indeterminate form \( \frac{0}{0} \). This is a situation where l'Hôpital's Rule can be applied.
3. Apply l'Hôpital's Rule, which states that if \( \lim_{x \to a} \frac{f(x)}{g(x)} \) 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 limit on the right exists.
4. Differentiate the numerator and the denominator separately: the derivative of \( x \) is \( 1 \), and the derivative of \( \sin x \) is \( \cos x \).
5. Now, evaluate the new limit: \( \lim_{x \to 0} \frac{1}{\cos x} \). Since \( \cos 0 = 1 \), the limit simplifies to \( \frac{1}{1} \).