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Magazine Chapter 27: Problem 27
Evaluate each limit (if it exists). Use \(L\) Hospital's rule (if appropriate). $$\lim _{x \rightarrow 0} \frac{2 x+\sin 3 x}{x-\sin 3 x}$$
Short Answer
Expert verified
The limit is \(-\frac{5}{2}\).
Step by step solution
01
Check the Form of the Limit
First, substitute \( x = 0 \) into the function to see if the limit can be directly evaluated or if \( L' \)H么pital's Rule might be needed. Substitute to get \[ \frac{2(0) + \sin(3 \times 0)}{0 - \sin(3 \times 0)} = \frac{0}{0}. \] Since this results in an indeterminate form \( \frac{0}{0} \), we may use \( L' \)H么pital's Rule.
02
Apply L'H么pital's Rule
L'H么pital's Rule states that if the limit results in the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then \[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \] if this limit exists. Here, differentiate the numerator and the denominator with respect to \( x \): * The derivative of the numerator \( 2x + \sin 3x \) is \( 2 + 3\cos 3x \). * The derivative of the denominator \( x - \sin 3x \) is \( 1 - 3\cos 3x \). Thus, our new limit is: \[ \lim_{x \to 0} \frac{2 + 3\cos 3x}{1 - 3\cos 3x}. \]
03
Evaluate the New Limit
Substitute \( x = 0 \) into the differentiated function: \[ \frac{2 + 3\cos(3 \times 0)}{1 - 3\cos(3 \times 0)} = \frac{2 + 3(1)}{1 - 3(1)} = \frac{2 + 3}{1 - 3} = \frac{5}{-2}. \] This simplified to \(-\frac{5}{2}\), which is not an indeterminate form. Thus, the limit exists.
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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 for tackling limits that present indeterminate forms. These forms include \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), where direct substitution doesn't lead to a clear answer.
When faced with such indeterminate forms, L'H么pital's Rule provides an alternative route:
When faced with such indeterminate forms, L'H么pital's Rule provides an alternative route:
- If \( \lim_{x \to a} \frac{f(x)}{g(x)} \) results in an indeterminate form, compute \( \lim_{x \to a} \frac{f'(x)}{g'(x)} \).
- The derivatives \( f'(x) \) and \( g'(x) \) represent the rates of change of the original functions.
- Use these derivatives to re-evaluate the limit.
Indeterminate Forms
Indeterminate forms arise in calculus when limits are difficult to evaluate directly. The most common types are \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \).
These forms signal that further analysis is needed since the straightforward substitution of \( x = a \) leads to unpredictable or undefined outcomes.
These forms signal that further analysis is needed since the straightforward substitution of \( x = a \) leads to unpredictable or undefined outcomes.
- \( \frac{0}{0} \) suggests that both the numerator and denominator approach zero simultaneously.
- \( \frac{\infty}{\infty} \) indicates both approach infinity.
- Identifying these forms is the first step towards solving limits effectively.
Differentiation
Differentiation is a core concept in calculus that measures how a function changes as its input changes. It's key to applying L'H么pital's Rule when assessing limits.
In differentiation, the derivative of a function \( f(x) \) at a point gives the slope of the tangent to the function's graph. It's a measure of the function's instantaneous rate of change at that point.
In differentiation, the derivative of a function \( f(x) \) at a point gives the slope of the tangent to the function's graph. It's a measure of the function's instantaneous rate of change at that point.
- The derivative \( f'(x) \) tells us how \( f(x) \) varies with a small change in \( x \).
- Techniques include the power rule, chain rule, and product rule to find derivatives.
- In the problem provided, differentiating "\( 2x + \sin 3x \)" yields "\( 2 + 3\cos 3x \)", capturing how changes occur around \( x = 0 \).
Evaluating Limits
In calculus, evaluating limits is essential for understanding how functions behave as they approach specific points. It involves finding the value that a function approaches as the input approaches a particular point.
A limit gives insight into a function's behavior near a point and can determine continuity and differentiability.
A limit gives insight into a function's behavior near a point and can determine continuity and differentiability.
- Substituting the point into the function is the first step in limit evaluation.
- If direct substitution leads to an indeterminate form, L'H么pital's Rule or different approaches are needed.
- For the given exercise, L'H么pital's Rule helped simplify the form to \(-\frac{5}{2}\), meaning the function behaves predictably as \( x \) approaches 0.
