Question

In Exercises \(25-34,\) find the limit. $$ \lim _{x \rightarrow 0^{+}} \frac{2}{\sin x} $$

Short answer

The limit of \( \frac{2}{\sin x} \) as \( x \) approaches zero from the positive side is 0.

Step-by-step solution

Analyze the function near \( x = 0 \) from the positive side

As \( x \) approaches zero from positive values, \( \sin x \) approaches zero as well. Since \( \sin x \) is in the denominator of the fraction, we are effectively dividing by values very close to zero, which makes the value of the fraction very large. Thus, the limit will be \( + \infty \). However, let's show this more formally.

Use L'Hopital's Rule

Since the function is of the form \( \frac{0}{0} \) as \( x \) approaches zero, we can apply l'Hopital's rule. This rule states that the limit of a quotient of two functions, where both approach zero or both approach infinity as \( x \) approaches a certain value, is equal to the limit of the quotients of their derivatives. The derivative of \( 2 \) is zero and the derivative of \( \sin x \) is \( \cos x \). Thus, we get \( \lim _{x \rightarrow 0^{+}} \frac{0}{\cos x} \) which is clearly zero.

Final Result

Therefore, \( \lim _{x \rightarrow 0^{+}} \frac{2}{\sin x} = 0 \).