Problem 15 - Thomas Calculus in SI Units

Question

Use l'Hôpital's rule to find the limits. $$\lim _{x \rightarrow 0} \frac{8 x^{2}}{\cos x-1}$$

Step-by-Step Solution

Answer

The limit is \(-16\).

Step 1: Identify the Indeterminate Form

Our task is to find the limit \( \lim_{x \rightarrow 0} \frac{8x^2}{\cos x - 1} \). First, we substitute \( x = 0 \) into the expression to check if it’s an indeterminate form. Both the numerator and the denominator result in zero:\[\text{Numerator: } 8(0)^2 = 0 \\text{Denominator: } \cos(0) - 1 = 1 - 1 = 0\]Thus, the expression is of the form \( \frac{0}{0} \), which is an indeterminate form suitable for l'Hôpital's rule.

Step 2: Apply l'Hôpital's Rule

Since the original expression is in the \( \frac{0}{0} \) indeterminate form, we can apply l'Hôpital's rule, which states to take the derivative of the numerator and the denominator separately:\[\lim_{x \rightarrow 0} \frac{8x^2}{\cos x - 1} = \lim_{x \rightarrow 0} \frac{\frac{d}{dx}(8x^2)}{\frac{d}{dx}(\cos x - 1)}\]The derivative of the numerator \( 8x^2 \) is \( 16x \), and the derivative of the denominator \( \cos x - 1 \) is \( -\sin x \).Thus, the limit becomes:\[\lim_{x \rightarrow 0} \frac{16x}{-\sin x}\]

Step 3: Re-evaluate the Limit

Now, substitute \( x = 0 \) into the new expression. \[\text{Numerator: } 16(0) = 0 \\text{Denominator: } -\sin(0) = 0\]This is again an indeterminate form \( \frac{0}{0} \), so we apply l'Hôpital's Rule once more to this new expression:\[\lim_{x \rightarrow 0} \frac{16x}{-\sin x} = \lim_{x \rightarrow 0} \frac{\frac{d}{dx}(16x)}{\frac{d}{dx}(-\sin x)}\]The derivative of \( 16x \) is 16, and the derivative of \( -\sin x \) is \( -\cos x \). Therefore, the limit becomes:\[\lim_{x \rightarrow 0} \frac{16}{-\cos x}\]

Step 4: Calculate the Final Limit

Now that the expression \( \frac{16}{-\cos x} \) no longer results in an indeterminate form, substitute \( x = 0 \):\[\lim_{x \rightarrow 0} \frac{16}{-\cos x} = \frac{16}{-\cos(0)} = \frac{16}{-1} = -16\]The limit is calculated to be \(-16\).

Key Concepts

Indeterminate Forms

When dealing with limits in calculus, we often encounter expressions that are difficult to evaluate directly. These are known as indeterminate forms. One common indeterminate form is \( \frac{0}{0} \), which means both the numerator and the denominator of a fraction approach zero as \( x \) approaches a certain value. This is what we saw in our original exercise.

• The fraction \( \frac{8x^2}{\cos x - 1} \) becomes zero over zero when \( x \to 0 \).
• This signals an opportunity to use l'Hôpital's rule, a technique designed to resolve indeterminate forms by differentiating.

Recognizing indeterminate forms is crucial because it indicates when l'Hôpital's rule can be applied. Without identifying such forms, using this powerful tool correctly would be challenging.

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a function is changing at any given point. In the context of l'Hôpital's rule, derivatives allow us to transform complex limits into simpler ones.

• In our example, the derivative of \( 8x^2 \) is \( 16x \).
• The derivative of \( \cos x - 1 \) is \( -\sin x \).
• We applied these derivatives to simplify the original limit expression.

By using derivatives, we turned the original indeterminate fraction into a new expression with the same limit behavior. The process typically involves computing derivatives until the indeterminate form is dissolved.

Limits

The concept of a limit is central to calculus and is used to define many other concepts, including the derivative itself. A limit seeks to find the value that a function approaches as its input approaches some point.

• In our exercise, we were calculating the limit \( \lim_{x \to 0} \frac{8x^2}{\cos x - 1} \).
• After applying derivatives, we resolved this to a simpler limit \( \lim_{x \to 0} \frac{16}{-\cos x} \), which could be evaluated directly.

Understanding limits involves both computation and intuition about function behavior. In many cases, using limits requires altering how the expression is approached through algebraic manipulation or calculus techniques like l'Hôpital's rule.

Calculus Exercises

Calculus exercises such as the one involving l'Hôpital's rule serve as both practice and application of calculus concepts. These exercises help reinforce our understanding and ability to handle different types of mathematical problems.

• Such problems typically involve identifying the right techniques or rules, like recognizing when l'Hôpital's rule applies.
• They often require applying basic calculus operations, such as finding derivatives and computing limits.

By engaging in these exercises, students not only learn how calculus works in theoretical settings but also how it can be applied to solve real-world problems. Mastery comes from practice and repeated exposure to similar problems, gradually building confidence and proficiency.