Question

Evaluate the limits with either L'Hôpital's rule or previously learned methods. $$ \lim _{x \rightarrow \pi} \frac{x-\pi}{\sin x} $$

Short answer

The limit is -1.

Step-by-step solution

Recognize the Indeterminate Form

First, identify that when we substitute \(x = \pi\) into \(\frac{x-\pi}{\sin x}\), we get \(\frac{0}{0}\). This is an indeterminate form which allows us to apply L'Hôpital's Rule.

Apply L'Hôpital's Rule

According to L'Hôpital's Rule, for limits of the form \(\frac{0}{0}\), we can take the derivative of the numerator and the denominator separately. The derivative of \(x-\pi\) is 1, and the derivative of \(\sin x\) is \(\cos x\).

Substitute and Evaluate the Limit

Replace \(\frac{x-\pi}{\sin x}\) with \(\frac{1}{\cos x}\) after differentiation. Now, evaluate the limit: \(\lim_{x \rightarrow \pi} \frac{1}{\cos x}\). Substitute \(x = \pi\) into \(\cos x\) which gives \(\cos \pi = -1\).

Simplify the Result

After substituting, the expression becomes \(\frac{1}{-1} = -1\). Therefore, the final result of the limit is -1.

Key concepts

L'Hôpital's Rule

When dealing with limits, especially those that result in indeterminate forms, L'Hôpital's Rule can be a lifesaver. This rule helps us simplify limits that initially appear as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), both classic examples of indeterminate forms.

Here's how L'Hôpital's Rule works: If the limit of a function as \(x\) approaches a value results in one of these indeterminate forms, you can take the derivative of the numerator and the derivative of the denominator separately. Then, calculate the limit of the new fraction.

In our original exercise, when \(x\) approaches \(\pi\), \(\frac{x-\pi}{\sin x}\) becomes \(\frac{0}{0}\). With L'Hôpital's Rule, we differentiate both parts to transform it into \(\frac{1}{\cos x}\). This allows us to easily find the limit without complicated algebraic manipulations.

Indeterminate Forms

In calculus, and particularly in the study of limits, indeterminate forms often arise. These are expressions where direct substitution leads to ambiguity, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) during limit evaluation.

Indeterminate forms signal that the limit cannot be determined without further manipulation or simplification. They are not defined because the value could actually approach different numbers, depending on the specific behavior of the function involved.

Importantly, not every fraction resembling an indeterminate form can be transformed with L'Hôpital’s Rule. It’s crucial that we check if these conditions are met. Other manipulation methods include factoring, dividing, or multiplying by the conjugate. These alternative strategies still rely heavily on identifying \(\frac{0}{0}\)-type situations.

Trigonometric Limits

Trigonometric functions, such as \(\sin(x)\) or \(\cos(x)\), frequently appear in limits equations, especially leading to indeterminate forms. Understanding their behavior around key angles (like \(0\), \(\frac{\pi}{2}\), \(\pi\)) is essential for evaluating such limits.

Some common strategies for handling limits involving trigonometric functions include using small-angle approximations or L'Hôpital's Rule. For example, limits like \(\lim_{x \to 0} \frac{\sin x}{x} = 1\) are classical results that aid in direct simplification of expressions.

In our problem, transforming \(\frac{x-\pi}{\sin x}\) to \(\frac{1}{\cos x}\) relied on knowledge of the derivative \(\cos x\), a concept often encountered when dealing with trigonometric functions in calculus limits.