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Magazine Chapter 2: Problem 29
Find the limits. $$\lim _{x \rightarrow 0} \frac{x+x \cos x}{\sin x \cos x}$$
Short Answer
Expert verified
The limit is 2.
Step by step solution
01
Express in Simplest Form
Firstly, write the given expression \( \frac{x + x \cos x}{\sin x \cos x} \) in a simplified form by factoring out \( x \) from the numerator. The expression becomes \( \frac{x(1 + \cos x)}{\sin x \cos x} \).
02
Apply Limit to Simplified Expression
Now, apply the limit to the simplified expression: \( \lim_{x \rightarrow 0} \frac{x(1 + \cos x)}{\sin x \cos x} \). This involves taking the limit of the individual components: numerators and denominators.
03
Analyze Limit at Zero
At \( x = 0 \), \( \sin x \rightarrow 0 \) and \( \cos x \rightarrow 1 \). Thus \( \cos x \rightarrow 1 \) creates no division by zero issue, but \( \sin x \cos x \rightarrow 0 \) challenges us to deal with indeterminacy. We proceed to apply L'H么pital's Rule to resolve it.
04
Use L'H么pital's Rule
Apply L'H么pital's Rule, which is suitable when the limit results in a \( \frac{0}{0} \) form. Differentiate the numerator and the denominator: the derivative of \( x(1 +\cos x) \) is \( 1 + \cos x - x \sin x \), and the derivative of \( \sin x \cos x \) is \( \cos^2 x - \sin^2 x \).
05
Evaluate the Differentiated Limit
Rewrite the differentiated expression: \( \lim_{x \rightarrow 0} \frac{1 + \cos x - x \sin x}{\cos^2 x - \sin^2 x} \). At \( x = 0 \), substituting gives \( \frac{1 + 1 - 0}{1 - 0} = 2 \).
06
Conclude the Limit
Since applying L'H么pital's Rule gives a determinate form, the limit evaluates to a real number. Therefore, the limit is \( 2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
L'H么pital's Rule
L'H么pital's Rule is a powerful tool in calculus for finding limits of indeterminate forms. When you encounter a limit of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), L'H么pital's Rule can be applied.
This rule states that if the limits of the numerator and the denominator both approach 0 or both approach infinity, you can differentiate the numerator and the denominator separately.
In our example, \( \lim_{x \rightarrow 0} \frac{x(1 + \cos x)}{\sin x \cos x} \), both the numerator and the denominator approach 0 as \( x \rightarrow 0 \).
Thus, we applied L'H么pital's Rule by differentiating to resolve the \( \frac{0}{0} \) form and correctly found a limit of 2.
This rule states that if the limits of the numerator and the denominator both approach 0 or both approach infinity, you can differentiate the numerator and the denominator separately.
- Take the derivative of the numerator.
- Take the derivative of the denominator.
- Re-evaluate the limit with these derivatives.
In our example, \( \lim_{x \rightarrow 0} \frac{x(1 + \cos x)}{\sin x \cos x} \), both the numerator and the denominator approach 0 as \( x \rightarrow 0 \).
Thus, we applied L'H么pital's Rule by differentiating to resolve the \( \frac{0}{0} \) form and correctly found a limit of 2.
Indeterminate Forms
Indeterminate forms are special instances where simple substitution does not easily reveal a limit. These forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), \( \infty - \infty \), \( 0^0 \), \( 1^\infty \), and \( \infty^0 \).
In our problem of \( \lim_{x \rightarrow 0} \frac{x(1 + \cos x)}{\sin x \cos x} \), substituting \( x = 0 \) initially gives \( \frac{0}{0} \), one of the common indeterminate forms.
Being an indeterminate form suggests that there's no straightforward evaluation, hence the need for techniques like L'H么pital's Rule.
In our problem of \( \lim_{x \rightarrow 0} \frac{x(1 + \cos x)}{\sin x \cos x} \), substituting \( x = 0 \) initially gives \( \frac{0}{0} \), one of the common indeterminate forms.
Being an indeterminate form suggests that there's no straightforward evaluation, hence the need for techniques like L'H么pital's Rule.
- Recognizing when a function is leading you toward an indeterminate form is crucial.
- Utilize L'H么pital's Rule, algebraic manipulation, or trigonometric identities to simplify and resolve these forms.
Trigonometric Limits
Trigonometric limits often involve expressions with functions like \( \sin x \) and \( \cos x \). As \( x \) approaches zero, these functions have well-known behaviors:
\( \sin x \rightarrow x \) and \( \cos x \rightarrow 1 \).
In the example \( \lim_{x \rightarrow 0} \frac{x(1 + \cos x)}{\sin x \cos x} \), recognizing the values \( \sin x \rightarrow 0 \) and \( \cos x \rightarrow 1 \) allows us to simplify and manipulate the expression more effectively.
\( \sin x \rightarrow x \) and \( \cos x \rightarrow 1 \).
In the example \( \lim_{x \rightarrow 0} \frac{x(1 + \cos x)}{\sin x \cos x} \), recognizing the values \( \sin x \rightarrow 0 \) and \( \cos x \rightarrow 1 \) allows us to simplify and manipulate the expression more effectively.
- Using these fundamental limits often aids in working through indeterminate forms.
- Remember that trigonometric identities like \( \sin 2x = 2 \sin x \cos x \) can be useful simplification tools.
