Problem 372 - Calculus Volume 1

Question

For the following exercises, evaluate the limits with either L'Hôpital's rule or previously learned methods. $$ \lim _{x \rightarrow 1} \frac{x-1}{\sin x} $$

Step-by-Step Solution

Answer

The limit is approximately 1.8508.

Step 1: Determine the Form of the Limit

As we substitute the limit value into the expression, we get \( \lim_{x \rightarrow 1} \frac{x - 1}{\sin x} \). Substituting \( x = 1 \), we find that the form is \( \frac{0}{0} \), which is an indeterminate form, allowing us to use L'Hôpital's rule.

Step 2: Apply L'Hôpital's Rule

L'Hôpital's Rule states that for limits of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \) if the limit exists. First, find the derivatives, \( f'(x) = 1 \) for \( f(x) = x - 1 \), and \( g'(x) = \cos x \) for \( g(x) = \sin x \).

Step 3: Evaluate the Limit

With the derivatives \( f'(x) = 1 \) and \( g'(x) = \cos x \), the limit must be evaluated as \( \lim_{x \to 1} \frac{1}{\cos x} \). Substitute \( x = 1 \) directly into the expression to get \( \frac{1}{\cos 1} \).

Step 4: Final Calculation

Calculate \( \cos 1 \) using a calculator to find its approximate value. Thus, \( \lim_{x \to 1} \frac{x-1}{\sin x} = \frac{1}{\cos 1} \) is defined, which is approximately \( \frac{1}{0.5403} \approx 1.8508 \).

Key Concepts

Limits

Limits help us understand the behavior of functions as they approach specific points. In our exercise, we approach the point where \( x \) goes to 1, for the expression \( \frac{x-1}{\sin x} \). When calculating limits, substituting the value directly into the function may lead to situations such as \( \frac{0}{0} \), which is known as an indeterminate form.

Here's the general idea:

• If the result is straightforward (like a simple number), that’s typically the limit.
• If not, we may need special techniques like L'Hôpital's Rule to find the limit accurately.

By understanding limits, we can determine how functions behave and how they rise or fall as they approach certain values or infinity.

Indeterminate Forms

Indeterminate forms are expressions that are inherently undefined without additional analysis. The two most famous forms are \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \). These forms tell us that the expressions could potentially converge to different values, and we need to explore further.

In our problem, substituting \( x = 1 \) directly into \( \frac{x-1}{\sin x} \) results in \( \frac{0}{0} \). This signals an indeterminate form, which guides us to apply L'Hôpital's Rule. Recognizing these situations helps us decide the right strategies to solve the limits without ambiguity, giving us accurate results.

Derivatives

Derivatives give us the measure of how a function changes as its input changes. They are central in calculating limits through L'Hôpital's Rule, specifically when dealing with indeterminate forms.

In our example, to solve \( \frac{x-1}{\sin x} \) using L'Hôpital's Rule, we first find the derivatives of the numerator and the denominator functions:

• The derivative of \( x-1 \) is 1.
• The derivative of \( \sin x \) is \( \cos x \).

Substituting these derivatives back into the limit gives \( \frac{1}{\cos x} \), which can now be evaluated straightforwardly as \( x \) approaches 1. This clear application of derivatives allows us to resolve indeterminate forms efficiently.