Problem 62
Question
Verify that L'Hôpital's rule is of no help in finding the limit; then find the limit, if it exists, by some other method. $$\lim _{x \rightarrow+\infty} \frac{x(2+\sin x)}{x^{2}+1}$$
Step-by-Step Solution
Verified Answer
The limit is 0.
1Step 1: Verify Indeterminate Form
To apply L'Hôpital's Rule, we need to first check if the limit is an indeterminate form. With our expression \( \frac{x(2+\sin x)}{x^2+1} \), as \( x \to +\infty \), the numerator approaches infinity (since \( x \to +\infty \)) and the denominator also approaches infinity (since \( x^2 + 1 \to +\infty \)). This could be an \( \frac{\infty}{\infty} \) form, allowing for L'Hôpital's Rule.
2Step 2: Determine Suitability for L'Hôpital's Rule
Checking the derivative-based requirements for L'Hôpital's Rule involves differentiating the numerator and the denominator. This results in a complex expression beyond basic algebraic simplification. More importantly, any direct use of L'Hôpital's Rule does not simplify or resolve the limit adequately, as the oscillatory term \( \sin x \) makes finding derivatives cumbersome and inefficient.
3Step 3: Simplify the Expression
Instead of using L'Hôpital's Rule, we analyze the expression by simplifying. Dividing every term in the numerator and the denominator by \( x^2 \) yields \( \frac{\frac{x(2+\sin x)}{x^2}}{\frac{x^2 + 1}{x^2}} = \frac{\frac{1}{x}(2+\sin x)}{1 + \frac{1}{x^2}} \). Note the dominant behavior as \( x \to +\infty \).
4Step 4: Evaluate Simplified Form
Consider each component as \( x \to +\infty \). The terms \( \frac{1}{x} \) and \( \frac{1}{x^2} \) both approach zero. Hence, the simplified expression becomes: \( \frac{0(2+\sin x)}{1+0} = 0 \). Regardless of the oscillating nature of \( \sin x \), its influence vanishes due to the factor \( \frac{1}{x} \).
5Step 5: Conclusion of the Limit
The function approaches 0 as \( x \to +\infty \). Thus, the limit of \( \frac{x(2+\sin x)}{x^2 + 1} \) is \( 0 \). This demonstrates that L'Hôpital's Rule wasn't necessary; the limit was found through simplification.
Key Concepts
Indeterminate FormsLimits at InfinityLimit Simplification Techniques
Indeterminate Forms
Sometimes, when evaluating limits, you may encounter expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These are known as indeterminate forms. They do not immediately reveal what the limit is and require further analysis. L'Hôpital's Rule helps analyze limits involving these forms by using derivatives. However, it's crucial first to check if an expression truly is an indeterminate form before using the rule.
In the given exercise, as \( x \) tends to infinity, both the numerator \( x(2+\sin x) \) and the denominator \( x^2 + 1 \) approach infinity. This signals an \( \frac{\infty}{\infty} \) indeterminate form. While this would usually allow L'Hôpital's Rule to be applied, complex oscillating terms like \( \sin x \) can complicate matters, making alternative methods more effective.
In the given exercise, as \( x \) tends to infinity, both the numerator \( x(2+\sin x) \) and the denominator \( x^2 + 1 \) approach infinity. This signals an \( \frac{\infty}{\infty} \) indeterminate form. While this would usually allow L'Hôpital's Rule to be applied, complex oscillating terms like \( \sin x \) can complicate matters, making alternative methods more effective.
Limits at Infinity
Understanding limits at infinity is essential for evaluating how a function behaves as \( x \) becomes very large (positively or negatively). In such scenarios, analyzing which terms grow the fastest helps identify the function's dominant behavior.
For the example \( \lim_{x \to +\infty} \frac{x(2+\sin x)}{x^2+1} \), as \( x \to +\infty \), the polynomial terms \( x \) and \( x^2 \) take precedence over oscillatory or constant terms like \( \sin x \). This is because polynomial terms increase unboundedly, while oscillatory terms like \( \sin x \) remain bounded between \(-1\) and \(1\) regardless of \( x \).
Recognizing this difference allows one to simplify the limit using different techniques, aside from applying L'Hôpital's Rule.
For the example \( \lim_{x \to +\infty} \frac{x(2+\sin x)}{x^2+1} \), as \( x \to +\infty \), the polynomial terms \( x \) and \( x^2 \) take precedence over oscillatory or constant terms like \( \sin x \). This is because polynomial terms increase unboundedly, while oscillatory terms like \( \sin x \) remain bounded between \(-1\) and \(1\) regardless of \( x \).
Recognizing this difference allows one to simplify the limit using different techniques, aside from applying L'Hôpital's Rule.
Limit Simplification Techniques
When direct application of L'Hôpital's Rule proves cumbersome or ineffective, simplification techniques help assess limits. In our scenario, instead of dealing with complex derivatives, simplifying the expressions simplifies the process. This is often done by dividing all terms by the highest power of \( x \) present in the denominator.
Let's take the expression \( \frac{x(2+\sin x)}{x^2+1} \). By dividing every term by \( x^2 \), it simplifies to \( \frac{\frac{1}{x}(2+\sin x)}{1+\frac{1}{x^2}} \). As \( x \to +\infty \), \( \frac{1}{x} \to 0 \) and \( \frac{1}{x^2} \to 0 \), which simplifies the expression further to \( \frac{0(2+\sin x)}{1+0} \), equaling zero.
This method highlights how recognizing dominant terms and applying algebraic simplifications can efficiently determine limits.
Let's take the expression \( \frac{x(2+\sin x)}{x^2+1} \). By dividing every term by \( x^2 \), it simplifies to \( \frac{\frac{1}{x}(2+\sin x)}{1+\frac{1}{x^2}} \). As \( x \to +\infty \), \( \frac{1}{x} \to 0 \) and \( \frac{1}{x^2} \to 0 \), which simplifies the expression further to \( \frac{0(2+\sin x)}{1+0} \), equaling zero.
This method highlights how recognizing dominant terms and applying algebraic simplifications can efficiently determine limits.
Other exercises in this chapter
Problem 61
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